#Leetcode# 917. Reverse Only Letters

最新推荐文章于 2025-09-11 19:19:55 发布

转载最新推荐文章于 2025-09-11 19:19:55 发布 · 85 阅读

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原文链接：http://www.cnblogs.com/zlrrrr/p/10398052.html

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#数据结构与算法

本文介绍了一种算法，该算法接收一个字符串作为输入，并返回一个新的字符串，其中所有非字母字符保持不变，而所有字母字符的位置则被反转。通过两个向量分别存储原始字符串中的字母和非字母字符，然后重新构建字符串，实现对字符串中字母位置的反转。

https://leetcode.com/problems/reverse-only-letters/

Given a string S, return the "reversed" string where all characters that are not a letter stay in the same place, and all letters reverse their positions.

Example 1:

Input: "ab-cd"
Output: "dc-ba"

Example 2:

Input: "a-bC-dEf-ghIj"
Output: "j-Ih-gfE-dCba"

Example 3:

Input: "Test1ng-Leet=code-Q!"
Output: "Qedo1ct-eeLg=ntse-T!"

Note:

S.length <= 100
33 <= S[i].ASCIIcode <= 122
S doesn't contain \ or "

代码：

class Solution {
public:
    string reverseOnlyLetters(string S) {
        string ans = "";
        vector<char> v;
        int len = S.length();
        for(int i = len - 1; i >= 0; i --) {
            if(S[i] >= 'a' && S[i] <= 'z' || S[i] >= 'A' && S[i] <= 'Z')
                v.push_back(S[i]);
        }
        
        int rec = 0;
        for(int i = 0; i < len; i ++) {
            if(S[i] >= 'a' && S[i] <= 'z' || S[i] >= 'A' && S[i] <= 'Z') {
                char c = v[rec];
                ans += c;
                rec ++;
            }
            else ans += S[i];
        }
        
        return ans;
    }
};

转载于:https://www.cnblogs.com/zlrrrr/p/10398052.html