[LeetCode]Reverse Words in a String I II

Reverse Words in a String I

题目描述

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.

代码

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

public class Solution{ //考虑前导空格、words之间多个空格 public String reverseWords(String s) { if(s == null || s.length() == 0) return ""; //一个空格或多个空格split String[] strs = s.trim().split("\s+"); StringBuilder sb = new StringBuilder(); for(int i = strs.length - 1; i > 0; i--){ sb.append(strs[i] + " "); } sb.append(strs[0]); return sb.toString(); } }

还有一种方法就是逆序遍历字符串，使用两个指针追踪words的开始和结尾，当我们遍历到words的开始时，将words append到结果中。代码如下：

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

public class Solution{ public String (String s){ if(s == null || s.length() == 0) return ""; StringBuilder sb = new StringBuilder(); //j指针初始指向字符串最后一个字符的下一位置，在每一次迭代中j指向上一个遍历的空格 int大专栏  [LeetCode]Reverse Words in a String I II> j = s.length(); for(int i = s.length() - 1; i >= 0; i--){ //遍历到空格，更新j的指向，指向新的空格 if(s.charAt(i) == ' '){ j = i; }else if(i == 0 || s.charAt(i - 1) == ' '){//charAt(i - 1) == ' '说明i当前指向某个word的开头，可以获取到该word if(sb.length() != 0){ sb.append(" "); } sb.append(s.substring(i, j)); } //其他情况直接跳过 } return sb.toString(); } }

Reverse Words in a String II

题目描述

Similar to Question [Reverse Words in a String], but with the following constraints:
“The input string does not contain leading or trailing spaces and the words are always separated by a single space.”
Could you do it in-place without allocating extra space?

代码

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

//将整个字符串翻转，然后将每个word翻转 public void (char[] s){ reverse(s, 0, s.length - 1); //j指向每个word的开始 for(int i = 0, j = 0; i < s.length; i++){ if(i == s.length - 1 || s[i] == ' ' ){ reverse(s, j, i); j = i+1; } } } public void reverse(char[] s, int start, int end){ while(start < end){ char temp = s[end]; s[end] = s[start]; s[start] = temp; } }