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
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