[leetcode 151]Reverse Words in a String

1 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".

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.

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 2 思路

根据题目提醒的思路，1 什么是一个单词：一串没有空格的字符 2 考虑前面和后面的空格  3 考虑两个单词之间多个空格

首先区分单词 java有个函数spilt，解决问题1

干掉前面和后面的单词，java有个函数 trim，解决问题2

得到带有""的单词数组后，可以再遍历一遍，若是""，则不加入，解决问题3

最后从威到头组成一个新的字符串，输出即可。

3 代码

    public String reverseWords(String s){    
        s = s.trim();
        String[] words = s.split(" ");
        
        ArrayList<String> arrayList = new ArrayList<String>();
        for (String string : words) {
            if (!string.isEmpty()) {
                arrayList.add(string);
            }
        }
        if (arrayList.isEmpty()) {
            return "";
        }
        
        int len = arrayList.size();
        
        StringBuffer sb = new StringBuffer();
        for (int i = 0; i < len - 1; i++) {//少加一个，避免最后一个单词后面有空格
            sb.append(arrayList.get(len - i - 1));
            sb.append(" ");
        }
        sb.append(arrayList.get(0));
    
        return sb.toString();
    }