LeetCode #186 - Reverse Words in a String II

最新推荐文章于 2022-08-04 09:16:14 发布

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最新推荐文章于 2022-08-04 09:16:14 发布

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本文介绍了一种高效的字符串反转算法，特别适用于单词级别的反转操作。通过迭代和条件判断，该方法能够不使用额外空间就地反转字符串中的每个单词，适用于C++等编程语言。此算法对于理解和实现字符串处理功能至关重要。

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题目描述：

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

Example:

Input:  ["t","h","e"," ","s","k","y"," ","i","s"," ","b","l","u","e"]
Output: ["b","l","u","e"," ","i","s"," ","s","k","y"," ","t","h","e"]

Note:

A word is defined as a sequence of non-space characters.
The input string does not contain leading or trailing spaces.
The words are always separated by a single space.

Follow up: Could you do it in-place without allocating extra space?

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