557. 反转字符串中的单词 III

给定一个字符串，你需要反转字符串中每个单词的字符顺序，同时仍保留空格和单词的初始顺序。

示例 1:

输入: "Let's take LeetCode contest"
输出: "s'teL ekat edoCteeL tsetnoc"
注意：在字符串中，每个单词由单个空格分隔，并且字符串中不会有任何额外的空格。

//方法一：
class Solution {
    public String reverseWords(String s) {
        String words[] = s.split(" ");
        StringBuilder res=new StringBuilder();
        for (String word: words)
            res.append(new StringBuffer(word).reverse().toString() + " ");
        return res.toString().trim();
    }
}
//方法二：不使用系统函数，自己实现
public class Solution {
    public String reverseWords(String s) {
        String words[] = split(s);
        StringBuilder res=new StringBuilder();
        for (String word: words)
            res.append(reverse(word) + " ");
        return res.toString().trim();
    }
    public String[] split(String s) {
        ArrayList < String > words = new ArrayList < > ();
        StringBuilder word = new StringBuilder();
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == ' ') {
                words.add(word.toString());
                word = new StringBuilder();
            } else
                word.append( s.charAt(i));
        }
        words.add(word.toString());
        return words.toArray(new String[words.size()]);
    }
    public String reverse(String s) {
      StringBuilder res=new StringBuilder();
        for (int i = 0; i < s.length(); i++)
            res.insert(0,s.charAt(i));
        return res.toString();
    }
}