LEETCODE: Reverse Words in a String

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本文介绍了一种用于将输入字符串中的单词进行反转的算法实现。通过遍历字符串并忽略前导空格，该方法能有效处理包含多个连续空格的情况，并确保返回的字符串中单词间仅由单个空格分隔，且不含前导或尾随空格。

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

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

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.

class Solution {
public:
    void reverseWords(string &s) {
        // Skip the spaces on the left
        int pos = 0;
        for(; pos < s.length(); pos ++) {
            if(s[pos] != ' ')
                break;
        }

        string result;
        string word;
        for(; pos < s.length(); pos ++) {
            // If is space, check if we have word. Have, record it. Not have, skip.
            if(s[pos] == ' ') {
                if(word.length() != 0) {
                    if(!result.empty())
                        result = " " + result;
                    result = word + result;
                    word.clear();
                }
            }
            else {
                word += s[pos];
            }
        }
        
        if(word != "") {
            if(!result.empty())
                result = " " + result;
            result = word + result;
        }
        
        if(result.length() != 0) {
            if(result[0] == ' ') {
                s = result.substr(1);
                return;
            }
        }
        s = result;
    }
};