[leetcode]151. Reverse Words in a String@Java解题报告

https://leetcode.com/problems/reverse-words-in-a-string/tabs/description

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.

思路

这道题让我们翻转字符串中的单词，题目中给了我们写特别说明，如果单词之间遇到多个空格，只能返回一个，而且首尾不能有单词，并且对C语言程序员要求空间复杂度为O(1)，我是用Java写的，就不考虑这个问题了。

先整个字符串整体翻转一次，然后再分别翻转每一个单词（或者先分别翻转每一个单词，然后再整个字符串整体翻转一次），此时就能得到我们需要的结果了。代码如下：

package go.jacob.day729;/** * 151. Reverse Words in a String *  * @author Jacob * */public class Demo2 {public String reverseWords(String s) {if (s == null || s.length() < 2)return s;char[] arr = s.toCharArray();int left = 0, right = s.length() - 1;// 完全翻转reverse(arr,left,right);//翻转单个单词reverseWord(arr);//空格处理return cleanSpaces(arr);}private String cleanSpaces(char[] arr) {int n=arr.length;int i=0,j=0;while(j<n){while(j<n&&arr[j]==' ')j++;while(j<n&&arr[j]!=' '){arr[i++]=arr[j++];}while(j<n&&arr[j]==' ')j++;if(j<n)arr[i++]=' ';}return new String(arr).substring(0,i);}private void reverseWord(char[] arr) {int n=arr.length,i=0,j=0;while(j<n){while(i<j||i<n&&arr[i]==' ')i++;while(j<i||j<n&&arr[j]!=' ')j++;reverse(arr,i,j-1);}}//翻转left到right间的字母private void reverse(char[] arr, int left, int right) {while(left<right){char temp=arr[left];arr[left]=arr[right];arr[right]=temp;left++;right--;}}}