lintcode :旋转字符串

题目：

旋转字符串

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