186. Reverse Words in a String II

最新推荐文章于 2025-08-15 16:44:13 发布

原创最新推荐文章于 2025-08-15 16:44:13 发布 · 163 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#string #array #leetcode #two pointers

本文详细解析了如何使用C++实现字符串单词级别的反转，包括去除多余空格、保持单词间单个空格，以及避免首尾空格的方法。通过具体示例和代码片段，展示了多种反转字符串的算法实现，特别关注了在不同情况下如何正确处理空格。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given an input string, reverse the string word by word.

Example 1:

Input: "the sky is blue"
Output: "blue is sky the"

Example 2:

Input: "  hello world!  "
Output: "world! hello"
Explanation: Your reversed string should not contain leading or trailing spaces.

Example 3:

Input: "a good   example"
Output: "example good a"
Explanation: You need to reduce multiple spaces between two words to a single space in the reversed string.

Note:

A word is defined as a sequence of non-space characters.
Input string may contain leading or trailing spaces. However, your reversed string should not contain leading or trailing spaces.
You need to reduce multiple spaces between two words to a single space in the reversed string.

方法1:

思路：

和151. Reverse Words in a String 一样。但是由于不考虑空格可以直接按照先大翻转再小翻转。

参考：grandyang https://www.cnblogs.com/grandyang/p/5186294.html
这里的loop写法非常值得借鉴

易错点

loop很容易写坏
下面的写法就谜一样的错了

//["t","h","e"," ","s","k","y"," ","i","s"," ","b","l","u","e"]
//["e","u","l","b"," ","s","i"," ","y","k","s"," ","e","h","t"]
//  s               e

//["b","l","u","e"," ","s","i"," ","y","k","s"," ","e","h","t"]
//  s               e
//                      s       e

//["b","l","u","e"," ","i","s"," ","y","k","s"," ","e","h","t"]
//                      s       e
//                                  s           e

//["b","l","u","e"," ","i","s"," ","s","k","y"," ","e","h","t"]
//                                  s           e
//                                              s             e

class Solution {
public:
    void reverseWords(vector<char>& str) {
        reverse(str.begin(), str.end());
       
        for (int i = 0, j = 0; i < str.size(); i = j + 1){
            for (j = i; j < str.size(); j ++){
                if (str[j] == ' ') break;
            }
            reverse(str.begin() + i, str.begin() + j);
        }
        return;
    }
};

class Solution {
public:
    void reverseWords(vector<char>& str) {
        reverse(str.begin(), str.end());
        int start = 0, end = 0;
        int n = str.size();
        while (end < n && start < n){
            while (str[end] != ' '){
                end++;
            }
            //reverse(str.begin() + start, str.begin() + end);
    
            start = end + 1;
            end = start;
        }
        return;
    }
};

class Solution {
public:
    void reverseWords(vector<char>& str) {
        reverse(str, 0, str.size() - 1);
        int begin = -1, end = 0, n = str.size();
        for (int end = 0; end < n; end++) {
            if (begin == -1) {
                if (str[end] != ' ') begin = end;
            }
            // 这里不能是else，因为可能begin == end
            if (end == n - 1 || str[end + 1] == ' ') {
                reverse(str, begin, end);
                begin = -1;
            }
        }
        return;
    }
    void reverse(vector<char>& str, int left, int right) {
        while (left < right) {
            swap(str[left++], str[right--]);
        }
        return;
    }
};