【leetcode】151. Reverse Words in a String

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最新推荐文章于 2025-08-09 17:01:50 发布

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本文介绍了一种算法，用于反转字符串中的单词顺序，并去除多余的空格。提供了C++实现代码，包括如何在原地进行操作以达到O(1)的空间复杂度。

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题目：

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

For example,
Given s = "the sky is blue",
return "blue is sky the".

Update (2015-02-12):
For C programmers: Try to solve it in-place in O(1) space.

click to show clarification.

Clarification:

What constitutes a word?
A sequence of non-space characters constitutes a word.
Could the input string contain leading or trailing spaces?
Yes. However, your reversed string should not contain leading or trailing spaces.
How about multiple spaces between two words?
Reduce them to a single space in the reversed string.

c++代码：

class Solution {
public:
    void reverseOneWord(string &s, int start, int end){
        while(start<end){
            swap(s[start++],s[end--]);
        }
    }
    void reverseWords(string &s) {
        int i=0;  //快
        int j=0;  //慢
        int len=s.size();
        int word_count=0;
        while(i<len){
            while(i<len && s[i]== ' '){  //找到下一个不为空的字符串
                i++;
            }//while
            if(i>=len){
                break;
            }
            if(word_count>0){  //如果不是第一个单词，那么每个单词前都需要加一个空格
                s[j++]=' ';
            }
            int start=j;
            for(;i<len && s[i]!=' ';i++,j++){  //从i开始的单词，移动到从j开始
                s[j]=s[i];
            }
            reverseOneWord(s, start, j-1);  //翻转当前单词
            word_count++;
        }//while
        s.resize(j);  //去掉最后一个单词后面的空字符串
        reverseOneWord(s, 0, j-1);  //翻转整个单词
    }
};