186. Reverse Words in a String II

题目：

Given an input string, reverse the string word by word. A word is defined as a sequence of non-space characters.

The input string does not contain leading or trailing spaces and the words are always separated by a single space.

For example,
Given s = "the sky is blue",
return "blue is sky the".

Could you do it in-place without allocating extra space?

链接：  http://leetcode.com/problems/reverse-words-in-a-string-ii/

题解：

翻转单词II。这次给定了char array，我们就可以使用三步翻转法了。因为题目的条件很优惠，前后没有空格，单词间又只有一个空格，那我们可以省略很多边界条件的判定，先翻转整个数组，然后碰到单词就正序回来。

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {
    public void reverseWords(char[] s) {
        if(s == null || s.length == 0)
            return;
        reverse(s, 0, s.length - 1);
        int lo = 0;
        
        for(int i = 0; i <= s.length; i++) {
            if(i == s.length || s[i] == ' ') {
                reverse(s, lo, i - 1);
                lo = i + 1;
            }
        }
    }
    
    private void reverse(char[] s, int i, int j) {
        while(i < j)
            swap(s, i++, j--);
    }
    
    private void swap(char[] s, int i, int j) {
        char tmp = s[i];
        s[i] = s[j];
        s[j] = tmp;
    }
}

 

题外话:

同事要去南极玩，先飞到阿根廷，再到乌斯怀亚，然后坐科考船过去。科考船一套大概10500刀，机票另算，好爽啊。以后我也要好好去旅游。

 

二刷：

跟一刷基本一样

Java:

public class Solution {
    public void reverseWords(char[] s) {
        if (s == null || s.length < 2) return;
        int len = s.length;
        reverse(s, 0, len - 1);
        int lo = 0;
        for (int i = 0; i < len; i++) {
            if (s[i] == ' ') {
                reverse(s, lo, i - 1);
                lo = i + 1;
            }
        }
        reverse(s, lo, len - 1);
    }
    
    private void reverse(char[] s, int lo, int hi) {
        while (lo < hi) {
            char tmp = s[lo];
            s[lo] = s[hi];
            s[hi] = tmp;
            lo++;
            hi--;
        }
    }
}

 

Reference：

http://www.zhihu.com/question/19857821