题目：
给定一个字符串和一个偏移量，根据偏移量旋转字符串(从左向右旋转)
样例
对于字符串 "abcdefg".
offset=0 => "abcdefg"
offset=1 => "gabcdef"
offset=2 => "fgabcde"
offset=3 => "efgabcd"
挑战
在数组上原地旋转，使用O(1)的额外空间
解题：
这个题目和这一题很像，前部分逆序，后部分逆序，整体逆序，这里注意的是offeset会大于字符串长度的情况，所以要对offeset处理：offeset = offeset%len
Java程序：
public class Solution {
/**
* @param str: an array of char
* @param offset: an integer
* @return: nothing
*/
public void rotateString(char[] str, int offset) {
// write your code here
int left = 0;
int right = str.length-1;
if(str!=null && str.length!=0){
offset = offset%(right+1);
rotateStr(str,0,right - offset);
rotateStr(str,right - offset+1,right);
rotateStr(str,0,right);
}
}
public void rotateStr(char[]str,int left,int right){
char tmp;
while(left<right){
tmp = str[left];
str[left] = str[right];
str[right] = tmp;
left++;
right--;
}
}
}
总耗时: 831 ms
Python程序：
class Solution:
# @param s: a list of char
# @param offset: an integer
# @return: nothing
def rotateString(self, s, offset):
# write you code here
if s!=None and len(s)!=0:
left = 0
right = len(s) - 1
offset = offset%(right+1)
self.rotateStr(s,0,right - offset)
self.rotateStr(s,right - offset + 1,right)
self.rotateStr(s,0,right) def rotateStr(self,s,left,right):
while left<right:
tmp = s[left]
s[left] = s[right]
s[right] = tmp
left += 1
right -= 1
总耗时: 233 ms