151. Reverse Words in a String

最新推荐文章于 2025-09-10 22:21:01 发布

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最新推荐文章于 2025-09-10 22:21:01 发布

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文章标签： java runtime

原文链接：https://segmentfault.com/a/1190000018171196

本文介绍了一种算法，用于反转字符串中的单词，同时处理了多个空格和前后空格的问题，确保了反转后的字符串清晰整洁。

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

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

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.

Follow up: For C programmers, try to solve it in-place in O(1) space.

难度：medium

题目：给定字符串，反转字符串中的单词。
注意：单词指一组非空字符。输入的字符串可能包含前后空格。你需要在两个单词之间只保留一个空格。

思路：先反转，然后逐个单词反转。

Runtime: 9 ms, faster than 37.01% of Java online submissions for Reverse Words in a String.
Memory Usage: 38.8 MB, less than 100.00% of Java online submissions for Reverse Words in a String.

public class Solution {
    public String reverseWords(String s) {
        StringBuilder sb = new StringBuilder(" " + s + " ");
        sb.reverse();
        StringBuilder result = new StringBuilder();
        int begin = 0;
        for (int i = 0; i < sb.length() - 1; i++) {
            char c1 = sb.charAt(i);
            char c2 = sb.charAt(i + 1);
            if (c1 == ' ' && c2 != ' ') {
                begin = i;
            } else if (c1 != ' ' && c2 == ' ') {
                for (int j = i; j >= begin; j--) {
                    result.append(sb.charAt(j));
                }
            }
        }
        
        return result.toString().trim();
    }
}