Merge pull request #196 from LetMeFly666/410

添加了问题“410.分割数组的最大值”的代码和题解

Author: LetMeFly666

Codes/0410-split-array-largest-sum.cpp

@@ -0,0 +1,58 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2024-01-21 13:16:50
+ * @LastEditors: LetMeFly
+ * @LastEditTime: 2024-01-21 13:26:47
+ */
+#ifdef _WIN32
+#include "_[1,2]toVector.h"
+#endif
+
+class Solution {
+private:
+    bool check(vector<int>& nums, int k, int s) {
+        int cnt = 0;
+        for (int t : nums) {
+            if (t > s) {
+                return false;
+            }
+            if (t + cnt > s) {
+                k--;
+                cnt = 0;
+            }
+            cnt += t;
+        }
+        return --k >= 0;
+    }
+public:
+    int splitArray(vector<int>& nums, int k) {
+        int l = 0, r = accumulate(nums.begin(), nums.end(), 0) + 1;
+        while (l < r) {
+            int mid = (l + r) >> 1;
+            if (check(nums, k, mid)) {
+                r = mid;
+            }
+            else {
+                l = mid + 1;
+            }
+        }
+        return l;
+    }
+};
+
+#ifdef _WIN32
+/*
+[7,2,5,10,8] 2
+
+*/
+int main() {
+    string s;
+    int t;
+    while (cin >> s >> t) {
+        vector<int> v = stringToVector(s);
+        Solution sol;
+        cout << sol.splitArray(v, t) << endl;
+    }
+    return 0;
+}
+#endif

Codes/0410-split-array-largest-sum.py

@@ -0,0 +1,30 @@
+'''
+Author: LetMeFly
+Date: 2024-01-21 13:29:22
+LastEditors: LetMeFly
+LastEditTime: 2024-01-21 13:33:--
+'''
+from typing import List
+
+class Solution:
+    def check(self, k: int, s: int) -> bool:
+        cnt = 0
+        for t in self.nums:
+            if t > s:
+                return False
+            if cnt + t > s:
+                k -= 1
+                cnt = 0
+            cnt += t
+        return k - 1 >= 0
+
+    def splitArray(self, nums: List[int], k: int) -> int:
+        self.nums = nums
+        l, r = 0, sum(nums) + 1
+        while l < r:
+            mid = (l + r) >> 1
+            if self.check(k, mid):
+                r = mid
+            else:
+                l = mid + 1
+        return l

README.md

@@ -158,6 +158,7 @@
 |0402.移掉K位数字|中等|<a href="https://leetcode.cn/problems/remove-k-digits/" target="_blank">题目地址</a>|<a href="https://blog.tisfy.eu.org/2022/10/15/LeetCode%200402.%E7%A7%BB%E6%8E%89K%E4%BD%8D%E6%95%B0%E5%AD%97/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/127332829" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/remove-k-digits/solution/letmefly-402yi-diao-k-wei-shu-zi-by-tisf-6hh0/" target="_blank">LeetCode题解</a>|
 |0406.根据身高重建队列|中等|<a href="https://leetcode.cn/problems/queue-reconstruction-by-height/" target="_blank">题目地址</a>|<a href="https://blog.tisfy.eu.org/2022/10/17/LeetCode%200406.%E6%A0%B9%E6%8D%AE%E8%BA%AB%E9%AB%98%E9%87%8D%E5%BB%BA%E9%98%9F%E5%88%97/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/127359227" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/queue-reconstruction-by-height/solution/letmefly-406gen-ju-shen-gao-zhong-jian-d-yfzu/" target="_blank">LeetCode题解</a>|
 |0409.最长回文串|简单|<a href="https://leetcode.cn/problems/longest-palindrome/" target="_blank">题目地址</a>|<a href="https://blog.tisfy.eu.org/2022/10/17/LeetCode%200409.%E6%9C%80%E9%95%BF%E5%9B%9E%E6%96%87%E4%B8%B2/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/127359550" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/longest-palindrome/solution/letmefly-409zui-chang-hui-wen-chuan-by-t-xsid/" target="_blank">LeetCode题解</a>|
+|0410.分割数组的最大值|困难|<a href="https://leetcode.cn/problems/split-array-largest-sum/solutions/" target="_blank">题目地址</a>|<a href="https://blog.tisfy.eu.org/2024/01/21/LeetCode%200410.%E5%88%86%E5%89%B2%E6%95%B0%E7%BB%84%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/135728821" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/split-array-largest-sum/solutions/2613456/letmefly-410fen-ge-shu-zu-de-zui-da-zhi-agh2v/" target="_blank">LeetCode题解</a>|
 |0415.字符串相加|简单|<a href="https://leetcode.cn/problems/add-strings/solutions/" target="_blank">题目地址</a>|<a href="https://blog.tisfy.eu.org/2023/07/17/LeetCode%200415.%E5%AD%97%E7%AC%A6%E4%B8%B2%E7%9B%B8%E5%8A%A0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/131758488" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/add-strings/solutions/2346886/letmefly-415zi-fu-chuan-xiang-jia-mo-ni-r1ph3/" target="_blank">LeetCode题解</a>|
 |0421.数组中两个数的最大异或值|中等|<a href="https://leetcode.cn/problems/maximum-xor-of-two-numbers-in-an-array/" target="_blank">题目地址</a>|<a href="https://blog.tisfy.eu.org/2022/10/19/LeetCode%200421.%E6%95%B0%E7%BB%84%E4%B8%AD%E4%B8%A4%E4%B8%AA%E6%95%B0%E7%9A%84%E6%9C%80%E5%A4%A7%E5%BC%82%E6%88%96%E5%80%BC/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/127413219" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/maximum-xor-of-two-numbers-in-an-array/solution/letmefly-421shu-zu-zhong-liang-ge-shu-de-zaxe/" target="_blank">LeetCode题解</a>|
 |0445.两数相加II|中等|<a href="https://leetcode.cn/problems/add-two-numbers-ii/" target="_blank">题目地址</a>|<a href="https://blog.tisfy.eu.org/2022/10/14/LeetCode%200445.%E4%B8%A4%E6%95%B0%E7%9B%B8%E5%8A%A0II/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/127316269" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/add-two-numbers-ii/solution/letmefly-445liang-shu-xiang-jia-ii-by-ti-vbh5/" target="_blank">LeetCode题解</a>|

Solutions/LeetCode 0410.分割数组的最大值.md

@@ -0,0 +1,150 @@
+---
+title: 410.分割数组的最大值
+date: 2024-01-21 13:35:04
+tags: [题解, LeetCode, 困难, 贪心, 数组, 二分查找, 二分, 动态规划, 前缀和]
+---
+
+# 【LetMeFly】410.分割数组的最大值：二分
+
+力扣题目链接：[https://leetcode.cn/problems/split-array-largest-sum/](https://leetcode.cn/problems/split-array-largest-sum/)
+
+<p>给定一个非负整数数组 <code>nums</code> 和一个整数&nbsp;<code>m</code> ，你需要将这个数组分成&nbsp;<code>m</code><em>&nbsp;</em>个非空的连续子数组。</p>
+
+<p>设计一个算法使得这&nbsp;<code>m</code><em>&nbsp;</em>个子数组各自和的最大值最小。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>nums = [7,2,5,10,8], m = 2
+<strong>输出：</strong>18
+<strong>解释：</strong>
+一共有四种方法将 nums 分割为 2 个子数组。 
+其中最好的方式是将其分为 [7,2,5] 和 [10,8] 。
+因为此时这两个子数组各自的和的最大值为18，在所有情况中最小。</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>nums = [1,2,3,4,5], m = 2
+<strong>输出：</strong>9
+</pre>
+
+<p><strong>示例 3：</strong></p>
+
+<pre>
+<strong>输入：</strong>nums = [1,4,4], m = 3
+<strong>输出：</strong>4
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
+	<li><code>0 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
+	<li><code>1 &lt;= m &lt;= min(50, nums.length)</code></li>
+</ul>
+
+
+    
+## 方法一：二分
+
+写一个函数```check(s)```，返回```能否```将数组```nums```划分为```k```部分且每一部分的最大值不超过```s```。
+
+> 实现方法很简单，使用一个变量```cnt```来记录当前部分的元素和。
+>
+> 遍历数组，如果当前元素大于```s```，则必不可能成功划分，直接返回```false```。
+>
+> 若```cnt```加上当前元素超过了```s```，则将当前元素划分为一组（```k--```、```cnt = 0```）。
+>
+> ```cnt```加上当前元素。
+>
+> 遍历结束后，判断```k - 1```是否大于等于```0```。若是，则返回```true```，否则返回```false```。
+
+有了这样的函数，我们只需要在主函数中写一个二分，枚举值```mid```是否能通过```check```。
+
+> ```l```初始值为```0```，```r```初始值为“无穷大”（数组中所有元素之和再加一）。
+>
+> 当```l < r```时，令$mid = \lfloor \frac{l+r}{2} \rfloor$。
+>
+> 如果```check(mid)```通过了，则说明```mid```为一种合法分法，尝试更小的值能否成功划分（令```r = mid```）
+>
+> 否则，说明```mid```太小了，无法划分，尝试更大的值能否成功划分（令```l = mid + 1```）
+
+二分结束后，```l = r```，返回```l```即为答案。
+
++ 时间复杂度$O(len(nums)\times \log \sum nums)$
++ 空间复杂度$O(1)$
+
+### AC代码
+
+#### C++
+
+```cpp
+class Solution {
+private:
+    bool check(vector<int>& nums, int k, int s) {
+        int cnt = 0;
+        for (int t : nums) {
+            if (t > s) {
+                return false;
+            }
+            if (t + cnt > s) {
+                k--;
+                cnt = 0;
+            }
+            cnt += t;
+        }
+        return --k >= 0;
+    }
+public:
+    int splitArray(vector<int>& nums, int k) {
+        int l = 0, r = accumulate(nums.begin(), nums.end(), 0) + 1;
+        while (l < r) {
+            int mid = (l + r) >> 1;
+            if (check(nums, k, mid)) {
+                r = mid;
+            }
+            else {
+                l = mid + 1;
+            }
+        }
+        return l;
+    }
+};
+```
+
+#### Python
+
+```python
+# from typing import List
+
+class Solution:
+    def check(self, k: int, s: int) -> bool:
+        cnt = 0
+        for t in self.nums:
+            if t > s:
+                return False
+            if cnt + t > s:
+                k -= 1
+                cnt = 0
+            cnt += t
+        return k - 1 >= 0
+
+    def splitArray(self, nums: List[int], k: int) -> int:
+        self.nums = nums
+        l, r = 0, sum(nums) + 1
+        while l < r:
+            mid = (l + r) >> 1
+            if self.check(k, mid):
+                r = mid
+            else:
+                l = mid + 1
+        return l
+```
+
+> 同步发文于CSDN，原创不易，转载经作者同意后请附上[原文链接](https://blog.tisfy.eu.org/2024/01/21/LeetCode%200410.%E5%88%86%E5%89%B2%E6%95%B0%E7%BB%84%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC/)哦~
+> Tisfy：[https://letmefly.blog.csdn.net/article/details/135728821](https://letmefly.blog.csdn.net/article/details/135728821)