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update: 添加问题“2537.统计好子数组的数目”的代码和题解(#878)
Signed-off-by: LetMeFly666 <[email protected]>
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‎Codes/2537-count-the-number-of-good-subarrays.cpp‎

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/*
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* @Author: LetMeFly
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* @Date: 2025-04-16 19:45:53
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-04-16 20:25:23
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*/
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#if defined(_WIN32) || defined(__APPLE__)
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#include "_[1,2]toVector.h"
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#endif
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typedef long long ll;
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class Solution {
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public:
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ll countGood(vector<int>& nums, int k) {
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ll ans = 0;
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unordered_map<int, int> ma;
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ll now = 0;
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for (int l = 0, r = 0; r < nums.size(); r++) {
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now += ma[nums[r]]++;
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while (now >= k) {
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now -= --ma[nums[l++]];
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}
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ans += l;
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}
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return ans;
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}
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};

‎Codes/2537-count-the-number-of-good-subarrays.go‎

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/*
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* @Author: LetMeFly
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* @Date: 2025-04-16 20:50:11
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-04-16 20:56:36
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*/
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package main
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func countGood(nums []int, k int) (ans int64) {
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times := map[int]int64{}
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now := int64(0)
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for l, r := 0, 0; r < len(nums); r++ {
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now += times[nums[r]]
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times[nums[r]]++
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for now >= int64(k) {
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times[nums[l]]--
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now -= times[nums[l]]
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l++
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}
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ans += int64(l)
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}
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return
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}

‎Codes/2537-count-the-number-of-good-subarrays.java‎

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/*
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* @Author: LetMeFly
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* @Date: 2025-04-16 20:27:13
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-04-16 20:49:27
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*/
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import java.util.Map;
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import java.util.HashMap;
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// // CE
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// class Solution {
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// public long countGood(int[] nums, int k) {
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// long ans = 0, now = 0;
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// Map<Integer, Integer> ma = new HashMap<>();
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// for (int l = 0, r = 0; r < nums.length; r++) {
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// now += ma[nums[r]]++;
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// while (now >= k) {
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// now -= --ma[nums[l++]];
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// }
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// ans += now;
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// }
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// return ans;
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// }
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// }
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class Solution {
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public long countGood(int[] nums, int k) {
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long ans = 0, now = 0;
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Map<Integer, Integer> ma = new HashMap<>();
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for (int l = 0, r = 0; r < nums.length; r++) {
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now += ma.merge(nums[r], 1, Integer::sum) - 1;
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while (now >= k) {
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now -= ma.merge(nums[l++], -1, Integer::sum);
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}
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ans += l;
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}
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return ans;
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}
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}

‎Codes/2537-count-the-number-of-good-subarrays.py‎

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'''
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Author: LetMeFly
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Date: 2025-04-16 20:23:21
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LastEditors: LetMeFly.xyz
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LastEditTime: 2025-04-16 20:26:37
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'''
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from typing import List
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from collections import defaultdict
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class Solution:
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def countGood(self, nums: List[int], k: int) -> int:
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times = defaultdict(int)
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ans = l = now = 0
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for t in nums:
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now += times[t]
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times[t] += 1
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while now >= k:
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times[nums[l]] -= 1
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now -= times[nums[l]]
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l += 1
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ans += l
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return ans

‎README.md‎

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|2529.正整数和负整数的最大计数|简单|<a href="https://leetcode.cn/problems/maximum-count-of-positive-integer-and-negative-integer/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/04/09/LeetCode%202529.%E6%AD%A3%E6%95%B4%E6%95%B0%E5%92%8C%E8%B4%9F%E6%95%B4%E6%95%B0%E7%9A%84%E6%9C%80%E5%A4%A7%E8%AE%A1%E6%95%B0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/137554489" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/maximum-count-of-positive-integer-and-negative-integer/solutions/2731345/letmefly-2529zheng-zheng-shu-he-fu-zheng-p367/" target="_blank">LeetCode题解</a>|
770770
|2530.执行K次操作后的最大分数|中等|<a href="https://leetcode.cn/problems/maximal-score-after-applying-k-operations/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/10/18/LeetCode%202530.%E6%89%A7%E8%A1%8CK%E6%AC%A1%E6%93%8D%E4%BD%9C%E5%90%8E%E7%9A%84%E6%9C%80%E5%A4%A7%E5%88%86%E6%95%B0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/133899145" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/maximal-score-after-applying-k-operations/solutions/2487511/letmefly-2530zhi-xing-k-ci-cao-zuo-hou-d-s4pm/" target="_blank">LeetCode题解</a>|
771771
|2535.数组元素和与数字和的绝对差|简单|<a href="https://leetcode.cn/problems/difference-between-element-sum-and-digit-sum-of-an-array/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/09/26/LeetCode%202535.%E6%95%B0%E7%BB%84%E5%85%83%E7%B4%A0%E5%92%8C%E4%B8%8E%E6%95%B0%E5%AD%97%E5%92%8C%E7%9A%84%E7%BB%9D%E5%AF%B9%E5%B7%AE/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/142568318" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/difference-between-element-sum-and-digit-sum-of-an-array/solutions/2931399/letmefly-2535shu-zu-yuan-su-he-yu-shu-zi-lp8o/" target="_blank">LeetCode题解</a>|
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|2537.统计好子数组的数目|中等|<a href="https://leetcode.cn/problems/count-the-number-of-good-subarrays/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/04/16/LeetCode%202537.%E7%BB%9F%E8%AE%A1%E5%A5%BD%E5%AD%90%E6%95%B0%E7%BB%84%E7%9A%84%E6%95%B0%E7%9B%AE/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/147288778" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/count-the-number-of-good-subarrays/solutions/3653508/letmefly-2537tong-ji-hao-zi-shu-zu-de-sh-9fmo/" target="_blank">LeetCode题解</a>|
772773
|2544.交替数字和|简单|<a href="https://leetcode.cn/problems/alternating-digit-sum/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/07/12/LeetCode%202544.%E4%BA%A4%E6%9B%BF%E6%95%B0%E5%AD%97%E5%92%8C/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/131673485" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/alternating-digit-sum/solutions/2340450/letmefly-2544jiao-ti-shu-zi-he-by-tisfy-aa0f/" target="_blank">LeetCode题解</a>|
773774
|2545.根据第K场考试的分数排序|中等|<a href="https://leetcode.cn/problems/sort-the-students-by-their-kth-score/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/12/21/LeetCode%202545.%E6%A0%B9%E6%8D%AE%E7%AC%ACK%E5%9C%BA%E8%80%83%E8%AF%95%E7%9A%84%E5%88%86%E6%95%B0%E6%8E%92%E5%BA%8F/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/144635776" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/sort-the-students-by-their-kth-score/solutions/3026570/letmefly-2545gen-ju-di-k-chang-kao-shi-d-rp50/" target="_blank">LeetCode题解</a>|
774775
|2549.统计桌面上的不同数字|简单|<a href="https://leetcode.cn/problems/count-distinct-numbers-on-board/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/03/23/LeetCode%202549.%E7%BB%9F%E8%AE%A1%E6%A1%8C%E9%9D%A2%E4%B8%8A%E7%9A%84%E4%B8%8D%E5%90%8C%E6%95%B0%E5%AD%97/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/136963667" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/count-distinct-numbers-on-board/solutions/2703112/letmefly-2549tong-ji-zhuo-mian-shang-de-992xw/" target="_blank">LeetCode题解</a>|

‎Solutions/LeetCode 2537.统计好子数组的数目.md‎

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---
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title: 2537.统计好子数组的数目：滑动窗口(双指针)
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date: 2025-04-16 21:01:12
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tags: [题解, LeetCode, 中等, 数组, 哈希表, 滑动窗口, 双指针]
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categories: [题解, LeetCode]
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---
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# 【LetMeFly】2537.统计好子数组的数目：滑动窗口(双指针)
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力扣题目链接：[https://leetcode.cn/problems/count-the-number-of-good-subarrays/](https://leetcode.cn/problems/count-the-number-of-good-subarrays/)
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<p>给你一个整数数组 <code>nums</code>&nbsp;和一个整数 <code>k</code>&nbsp;，请你返回 <code>nums</code>&nbsp;中 <strong>好</strong>&nbsp;子数组的数目。</p>
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<p>一个子数组 <code>arr</code>&nbsp;如果有 <strong>至少</strong>&nbsp;<code>k</code>&nbsp;对下标 <code>(i, j)</code>&nbsp;满足 <code>i &lt; j</code>&nbsp;且 <code>arr[i] == arr[j]</code>&nbsp;，那么称它是一个 <strong>好</strong>&nbsp;子数组。</p>
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<p><strong>子数组</strong>&nbsp;是原数组中一段连续 <strong>非空</strong>&nbsp;的元素序列。</p>
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<p>&nbsp;</p>
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<p><strong>示例 1：</strong></p>
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<pre><b>输入：</b>nums = [1,1,1,1,1], k = 10
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<b>输出：</b>1
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<b>解释：</b>唯一的好子数组是这个数组本身。
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre><b>输入：</b>nums = [3,1,4,3,2,2,4], k = 2
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<b>输出：</b>4
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<b>解释：</b>总共有 4 个不同的好子数组：
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- [3,1,4,3,2,2] 有 2 对。
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- [3,1,4,3,2,2,4] 有 3 对。
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- [1,4,3,2,2,4] 有 2 对。
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- [4,3,2,2,4] 有 2 对。
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</pre>
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<p>&nbsp;</p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
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<li><code>1 &lt;= nums[i], k &lt;= 10<sup>9</sup></code></li>
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</ul>
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## 解题方法：滑动窗口(双指针)
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### 解题思路
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使用一个哈希表记录窗口(两个指针之间)中每个元素出现的次数。
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右指针从第一个元素到最后一个元素遍历数组，在移动的过程中，将指向的元素添加到“窗口”中并更新窗口中的“相等对数”；
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在窗口中的“相等对数”大于等于$k$时，不断右移左指针（将左指针指向元素移出窗口）并更新窗口中的“相等对数”；
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这样，右指针每次移动一个位置，左指针指向的就是第一个“窗口中相等对数小于$k$”的位置。
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### 具体方法
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假设右移**右**指针新**加入**窗口中了一个元素$x$，那么如何更新“窗口中的相等对数”呢？
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> **新加入**的$x$与之前窗口中的每个**已有**$x$都能配对，所以新增“对数”为新元素**加入前**哈希表中$x$的个数。
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假设右移**左**指针新**移除**了窗口中一个元素$x$，那么如何更新“窗口中的相等对数”呢？
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> **被移除**的$x$与窗口中的每个**其他**$x$都配过对，所以减少的“对数”为新元素**移除后**哈希表中$x$的个数。
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假设第一个“使窗口中相等对数小于$k$”的左指针下标为$l$，那么答案应该增加几？
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> $l$下标左边的任何一个元素开始，到右指针$r$的字数组都满足“相等对数大于等于$k$”，都是“好字数组”。
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>
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> 以$r$结尾的好字数组的个数为$l$左边的元素个数，对答案的贡献为$l$。
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### 时空复杂度
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+ 时间复杂度$O(len(nums))$。左右指针最多各自遍历数组一次
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+ 空间复杂度$O(1)$
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### AC代码
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#### C++
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```cpp
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/*
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* @Author: LetMeFly
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* @Date: 2025-04-16 19:45:53
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-04-16 20:25:23
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*/
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typedef long long ll;
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class Solution {
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public:
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ll countGood(vector<int>& nums, int k) {
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ll ans = 0;
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unordered_map<int, int> ma;
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ll now = 0;
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for (int l = 0, r = 0; r < nums.size(); r++) {
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now += ma[nums[r]]++;
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while (now >= k) {
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now -= --ma[nums[l++]];
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}
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ans += l;
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}
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return ans;
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}
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};
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```
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#### Python
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```python
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'''
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Author: LetMeFly
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Date: 2025-04-16 20:23:21
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LastEditors: LetMeFly.xyz
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LastEditTime: 2025-04-16 20:26:37
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'''
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from typing import List
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from collections import defaultdict
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class Solution:
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def countGood(self, nums: List[int], k: int) -> int:
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times = defaultdict(int)
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ans = l = now = 0
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for t in nums:
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now += times[t]
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times[t] += 1
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while now >= k:
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times[nums[l]] -= 1
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now -= times[nums[l]]
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l += 1
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ans += l
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return ans
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```
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#### Java
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```java
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/*
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* @Author: LetMeFly
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* @Date: 2025-04-16 20:27:13
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-04-16 20:49:27
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*/
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import java.util.Map;
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import java.util.HashMap;
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class Solution {
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public long countGood(int[] nums, int k) {
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long ans = 0, now = 0;
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Map<Integer, Integer> ma = new HashMap<>();
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for (int l = 0, r = 0; r < nums.length; r++) {
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now += ma.merge(nums[r], 1, Integer::sum) - 1;
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while (now >= k) {
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now -= ma.merge(nums[l++], -1, Integer::sum);
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}
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ans += l;
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}
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return ans;
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}
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}
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```
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#### Go
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```go
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/*
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* @Author: LetMeFly
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* @Date: 2025-04-16 20:50:11
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-04-16 20:56:36
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*/
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package main
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func countGood(nums []int, k int) (ans int64) {
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times := map[int]int64{}
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now := int64(0)
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for l, r := 0, 0; r < len(nums); r++ {
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now += times[nums[r]]
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times[nums[r]]++
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for now >= int64(k) {
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times[nums[l]]--
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now -= times[nums[l]]
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l++
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}
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ans += int64(l)
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}
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return
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}
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```
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> 同步发文于[CSDN](https://letmefly.blog.csdn.net/article/details/147288778)和我的[个人博客](https://blog.letmefly.xyz/)，原创不易，转载经作者同意后请附上[原文链接](https://blog.letmefly.xyz/2025/04/16/LeetCode%202537.%E7%BB%9F%E8%AE%A1%E5%A5%BD%E5%AD%90%E6%95%B0%E7%BB%84%E7%9A%84%E6%95%B0%E7%9B%AE/)哦~
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>
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> 千篇源码题解[已开源](https://github.com/LetMeFly666/LeetCode)

‎Solutions/Other-English-LearningNotes-SomeWords.md‎

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|voucher|n. 保证人，证件，教育补助券，消费券<br/>v. 证明，为...准备凭据，发给...凭证|
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|deterioration|n. 恶化，退化，变质，堕落|
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|||
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|wordy|adj. 话多的，冗长的，啰嗦的|
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|oceanography|n. 海洋学|
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|stronghold|n. 要塞，据点，堡垒，有广泛支持的地方|
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|mackintosh|n. 防水胶布，胶布雨衣（麦金托什）|
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|||
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|ferrous|adj. 含铁的，铁的|
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<p class="wordCounts">单词收录总数</p>
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‎changes20250407.patch‎

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