【Leetcode】Reverse word in a string

【题目】

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.

【题解】

1.遇到空格就跳过，第一个不是空格的元素，位置用pos记录。

2. 即pos之后，继续，只要不是空格，i++

3.当遇到的是第一个单词，即res长度为0时，直接添加，否则，添加“ ”，然后再在前面添加新的单词，达到reverse的效果，res= s.substring(pos,i),+res;注意顺序

4.i-- 因为在检查是否空格那里i++了

注意：substring函数，start是包含的，而end的标是达不到的；

用+重新赋值的好处是可以直接把新的单词调整到最前面，但是这样的做法是不是要耗费的空间比较大。得是单词数目+空格数目。基本是2倍单词数量。

【代码】

public static String solution(String s){String res= "";for(int i=0;i<s.length();i++){if(s.charAt(i)==' ') continue;int pos=i;while(i<s.length()&&s.charAt(i)!=' ') i++;if(res.length()>0) res=" "+res;res=(s.substring(pos,i))+res;i--;}return res;}

【官方题解】

One simple approach is a two-pass solution: First pass to split the string by spaces into an array of words, then second pass to extract the words in reversed order.

We can do better in one-pass. While iterating the string in reverse order, we keep track of a word’s begin and end position. When we are at the beginning of a word, we append it.

public class Solution {    public String reverseWords(String s) {    String[] a=s.trim().split("\\s+");StringBuilder res=new StringBuilder();for(int i=a.length-1;i>=0;i--){if(i==a.length-1)res.append(a[i]);else res.append(" ").append(a[i]);}return res.toString();    }}

\\s+:

splitpublic String[] split(String regex)根据给定的正则表达式的匹配来拆分此字符串。 然后就要明确正则表达式的含义了：\\s表示   空格,回车,换行等空白符,     +号表示一个或多个的意思,所以...