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
View Code
View Code
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:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
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:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
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