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update: 添加问题“386.字典序排数”的代码和题解(#975)
* MGJW(words): archive (#974) Signed-off-by: LetMeFly666 <814114971@qq.com> * 386: WA.cpp (#974) Signed-off-by: LetMeFly666 <814114971@qq.com> * 386: AC.cpp+WA.py (#974) cpp - AC,100.00%,87.60% Signed-off-by: LetMeFly666 <814114971@qq.com> * 386: AC.py (#974) while now % 10 == 9和if now == n不可以分开 否则now==n后now/=10后now%10还是可能等于9 py - AC,97.55%,98.68% Signed-off-by: LetMeFly666 <814114971@qq.com> * 386+MGJW: CE.java (#974) Signed-off-by: LetMeFly666 <814114971@qq.com> * 386: CE.java (#974) Signed-off-by: LetMeFly666 <814114971@qq.com> * 386: AC.java+go (#974) java - AC,83.25%,83.25% go - AC,100.00%,89.92% Signed-off-by: LetMeFly666 <814114971@qq.com> * update: 添加问题“386.字典序排数”的代码和题解(#975) Signed-off-by: LetMeFly666 <814114971@qq.com> * fix: Update Solutions/LeetCode 0386.字典序排数.md (#864) Co-authored-by: Copilot <175728472+Copilot@users.noreply.github.com> * fix: Update Solutions/LeetCode 0386.字典序排数.md (#864) Co-authored-by: Copilot <175728472+Copilot@users.noreply.github.com> --------- Signed-off-by: LetMeFly666 <814114971@qq.com> Co-authored-by: Copilot <175728472+Copilot@users.noreply.github.com>
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‎Codes/0386-lexicographical-numbers.cpp‎

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/*
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* @Author: LetMeFly
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* @Date: 2025-06-09 10:09:21
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-06-09 10:15:25
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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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class Solution {
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public:
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vector<int> lexicalOrder(int n) {
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vector<int> ans(n);
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int now = 1;
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for (int i = 0; i < n; i++) {
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ans[i] = now;
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if (now * 10 <= n) {
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now *= 10;
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} else {
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while (now % 10 == 9 || now == n) {
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now /= 10;
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}
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now++;
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}
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}
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return ans;
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}
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};

‎Codes/0386-lexicographical-numbers.go‎

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/*
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* @Author: LetMeFly
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* @Date: 2025-06-09 10:09:21
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-06-09 22:03:40
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*/
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package main
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func lexicalOrder(n int) []int {
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ans := make([]int, n)
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for now, i := 1, 0; i < n; i++ {
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ans[i] = now
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if now * 10 <= n {
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now *= 10
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} else {
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for now % 10 == 9 || now == n {
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now /= 10
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}
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now++
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}
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}
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return ans
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}

‎Codes/0386-lexicographical-numbers.java‎

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/*
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* @Author: LetMeFly
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* @Date: 2025-06-09 10:09:21
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-06-09 22:00:48
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*/
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import java.util.List;
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import java.util.ArrayList;
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class Solution {
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public List<Integer> lexicalOrder(int n) {
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List<Integer> ans = new ArrayList<>();
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for (int now = 1, i = 0; i < n; i++) {
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ans.add(now);
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if (now * 10 <= n) {
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now *= 10;
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} else {
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while (now % 10 == 9 || now == n) {
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now /= 10;
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}
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now++;
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}
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}
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return ans;
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}
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}

‎Codes/0386-lexicographical-numbers.py‎

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'''
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Author: LetMeFly
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Date: 2025-06-09 10:09:21
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LastEditors: LetMeFly.xyz
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LastEditTime: 2025-06-09 10:21:53
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'''
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from typing import List
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class Solution:
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def lexicalOrder(self, n: int) -> List[int]:
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ans = [0] * n
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now = 1
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for i in range(n):
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ans[i] = now
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if now * 10 <= n:
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now *= 10
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else:
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while now % 10 == 9 or now == n:
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now //= 10
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now += 1
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return ans

‎README.md‎

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|0377.组合总和Ⅳ|中等|<a href="https://leetcode.cn/problems/combination-sum-iv/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/10/10/LeetCode%200377.%E7%BB%84%E5%90%88%E6%80%BB%E5%92%8C%E2%85%A3/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/127251669" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/combination-sum-iv/solution/letmefly-377zu-he-zong-he-iv-by-tisfy-zp8z/" target="_blank">LeetCode题解</a>|
289289
|0381.O(1)时间插入、删除和获取随机元素-允许重复|困难|<a href="https://leetcode.cn/problems/insert-delete-getrandom-o1-duplicates-allowed/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/10/11/LeetCode%200381.O%281%29%E6%97%B6%E9%97%B4%E6%8F%92%E5%85%A5%E3%80%81%E5%88%A0%E9%99%A4%E5%92%8C%E8%8E%B7%E5%8F%96%E9%9A%8F%E6%9C%BA%E5%85%83%E7%B4%A0-%E5%85%81%E8%AE%B8%E9%87%8D%E5%A4%8D/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/127262069" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/insert-delete-getrandom-o1-duplicates-allowed/solution/letmefly-stlde-ying-yong-381o1-shi-jian-6ngis/" target="_blank">LeetCode题解</a>|
290290
|0383.赎金信|简单|<a href="https://leetcode.cn/problems/ransom-note/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/01/07/LeetCode%200383.%E8%B5%8E%E9%87%91%E4%BF%A1/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/135438384" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/ransom-note/solutions/2594460/letmefly-383shu-jin-xin-ji-shu-by-tisfy-lfty/" target="_blank">LeetCode题解</a>|
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|0386.字典序排数|中等|<a href="https://leetcode.cn/problems/lexicographical-numbers/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/06/09/LeetCode%200386.%E5%AD%97%E5%85%B8%E5%BA%8F%E6%8E%92%E6%95%B0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/148545551" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/lexicographical-numbers/solutions/3696853/letmefly-386zi-dian-xu-pai-shu-xi-xin-zo-48zm/" target="_blank">LeetCode题解</a>|
291292
|0387.字符串中的第一个唯一字符|简单|<a href="https://leetcode.cn/problems/first-unique-character-in-a-string/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/10/11/LeetCode%200387.%E5%AD%97%E7%AC%A6%E4%B8%B2%E4%B8%AD%E7%9A%84%E7%AC%AC%E4%B8%80%E4%B8%AA%E5%94%AF%E4%B8%80%E5%AD%97%E7%AC%A6/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/127262237" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/first-unique-character-in-a-string/solution/letmefly-387zi-fu-chuan-zhong-de-di-yi-g-gk9b/" target="_blank">LeetCode题解</a>|
292293
|0392.判断子序列|简单|<a href="https://leetcode.cn/problems/is-subsequence/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/10/12/LeetCode%200392.%E5%88%A4%E6%96%AD%E5%AD%90%E5%BA%8F%E5%88%97/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/127279017" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/is-subsequence/solution/letmefly-392pan-duan-zi-xu-lie-by-tisfy-o9f1/" target="_blank">LeetCode题解</a>|
293294
|0395.至少有K个重复字符的最长子串|中等|<a href="https://leetcode.cn/problems/longest-substring-with-at-least-k-repeating-characters/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/10/13/LeetCode%200395.%E8%87%B3%E5%B0%91%E6%9C%89K%E4%B8%AA%E9%87%8D%E5%A4%8D%E5%AD%97%E7%AC%A6%E7%9A%84%E6%9C%80%E9%95%BF%E5%AD%90%E4%B8%B2/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/127296624" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/longest-substring-with-at-least-k-repeating-characters/solution/by-tisfy-fv9c/" target="_blank">LeetCode题解</a>|

‎Solutions/LeetCode 0386.字典序排数.md‎

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---
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title: 386.字典序排数：细心总结条件
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date: 2025-06-09 23:36:33
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tags: [题解, LeetCode, 中等, 深度优先搜索, 字典树]
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categories: [题解, LeetCode]
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---
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# 【LetMeFly】386.字典序排数：细心总结条件
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力扣题目链接：[https://leetcode.cn/problems/lexicographical-numbers/](https://leetcode.cn/problems/lexicographical-numbers/)
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<p>给你一个整数 <code>n</code> ，按字典序返回范围 <code>[1, n]</code> 内所有整数。</p>
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<p>你必须设计一个时间复杂度为 <code>O(n)</code> 且使用 <code>O(1)</code> 额外空间的算法。</p>
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<p>&nbsp;</p>
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<p><strong>示例 1：</strong></p>
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<pre>
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<strong>输入：</strong>n = 13
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<strong>输出：</strong>[1,10,11,12,13,2,3,4,5,6,7,8,9]
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre>
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<strong>输入：</strong>n = 2
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<strong>输出：</strong>[1,2]
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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;= n &lt;= 5 * 10<sup>4</sup></code></li>
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</ul>
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## 解题方法：if-else
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### 解题思路
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使用O(1)的空间和O(n)的时间完成本题，不如先想想字典序下到底是个什么顺序。
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1. 首先是1、10、100、...，直到即将大于n为止（假设n=200，那么100即将大于n(100*10=1000)）
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2. 100后接着是101、102、...109
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3. 109之后是11（而不是110），之后进入第一步(11、110、...)
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基本上就这三种情况。
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### 具体做法
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使用一个变量now记录当前的值，如果$now\times 10\leq n$就让$now\times 10$，否则：
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> 就让$now+1$。特别的，在$now+1$之前，若$now$的最后一位已经是$9$或者$now$已经等于了$n$，就不断移除$now$的最后一位(例如，109变成10)。
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### 时空复杂度分析
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+ 时间复杂度$O(n)$
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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: 2025-06-09 10:09:21
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-06-09 10:15:25
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*/
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class Solution {
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public:
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vector<int> lexicalOrder(int n) {
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vector<int> ans(n);
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int now = 1;
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for (int i = 0; i < n; i++) {
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ans[i] = now;
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if (now * 10 <= n) {
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now *= 10;
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} else {
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while (now % 10 == 9 || now == n) {
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now /= 10;
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}
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now++;
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}
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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: 2025-06-09 10:09:21
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LastEditors: LetMeFly.xyz
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LastEditTime: 2025-06-09 10:21:53
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'''
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from typing import List
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class Solution:
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def lexicalOrder(self, n: int) -> List[int]:
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ans = [0] * n
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now = 1
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for i in range(n):
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ans[i] = now
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if now * 10 <= n:
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now *= 10
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else:
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while now % 10 == 9 or now == n:
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now //= 10
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now += 1
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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: 2025-06-09 10:09:21
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-06-09 22:00:48
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*/
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import java.util.List;
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import java.util.ArrayList;
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class Solution {
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public List<Integer> lexicalOrder(int n) {
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List<Integer> ans = new ArrayList<>();
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for (int now = 1, i = 0; i < n; i++) {
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ans.add(now);
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if (now * 10 <= n) {
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now *= 10;
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} else {
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while (now % 10 == 9 || now == n) {
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now /= 10;
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}
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now++;
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}
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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: 2025-06-09 10:09:21
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-06-09 22:03:40
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*/
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package main
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func lexicalOrder(n int) []int {
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ans := make([]int, n)
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for now, i := 1, 0; i < n; i++ {
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ans[i] = now
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if now * 10 <= n {
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now *= 10
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} else {
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for now % 10 == 9 || now == n {
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now /= 10
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}
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now++
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}
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}
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return ans
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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/148545551)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/2025/06/09/LeetCode%200386.%E5%AD%97%E5%85%B8%E5%BA%8F%E6%8E%92%E6%95%B0/)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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|||
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|stupidity|n. 愚蠢，愚蠢行为，糊涂，笨|
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|mutton|n. 羊肉|
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|||
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|plume|n. 羽毛，飘升之物，一缕，(常用作装饰的)连在一起的羽毛<br/>v. 用羽毛装饰，(鸟)整理(羽毛)，借衣服修饰，自夸|
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|indefinitely|adv. 无期限的|
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<p class="wordCounts">单词收录总数</p>
12101214

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