共计 1016 个字符，预计需要花费 3 分钟才能阅读完成。
Given an input string, reverse the string word by word.Example:
Input: “the sky is blue”,
Output: “blue is sky the”.
Note:
A word is defined as a sequence of non-space characters.
Input string may contain leading or trailing spaces. However, your reversed string should not contain leading or trailing spaces.
You need to reduce multiple spaces between two words to a single space in the reversed string.
Follow up: For C programmers, try to solve it in-place in O(1) space.
难度：medium
题目：给定字符串，反转字符串中的单词。注意：单词指一组非空字符。输入的字符串可能包含前后空格。你需要在两个单词之间只保留一个空格。
思路：先反转，然后逐个单词反转。
Runtime: 9 ms, faster than 37.01% of Java online submissions for Reverse Words in a String.Memory Usage: 38.8 MB, less than 100.00% of Java online submissions for Reverse Words in a String.
public class Solution {
public String reverseWords(String s) {
StringBuilder sb = new StringBuilder(” ” + s + ” “);
sb.reverse();
StringBuilder result = new StringBuilder();
int begin = 0;
for (int i = 0; i < sb.length() – 1; i++) {
char c1 = sb.charAt(i);
char c2 = sb.charAt(i + 1);
if (c1 == ‘ ‘ && c2 != ‘ ‘) {
begin = i;
} else if (c1 != ‘ ‘ && c2 == ‘ ‘) {
for (int j = i; j >= begin; j–) {
result.append(sb.charAt(j));
}
}
}
return result.toString().trim();
}
}
