# Rotate String (LintCode)

https://www.lintcode.com/problem/rotate-string/description

### Description

Given a string and an offset, rotate string by offset. (rotate from left to right)

Have you met this question in a real interview?

Yes

Problem Correction

### Example

Given`"abcdefg"`.

``````offset=0 => "abcdefg"
offset=1 => "gabcdef"
offset=2 => "fgabcde"
offset=3 => "efgabcd"
``````

### Challenge

Rotate in-place with O(1) extra memory.

## Solution

``````/**
* This reference program is provided by @jiuzhang.com
* Copyright is reserved. Please indicate the source for forwarding
*/

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
if (str == null || str.length == 0)
return;

offset = offset % str.length;
reverse(str, 0, str.length - offset - 1);
reverse(str, str.length - offset, str.length - 1);
reverse(str, 0, str.length - 1);
}

private void reverse(char[] str, int start, int end) {
for (int i = start, j = end; i < j; i++, j--) {
char temp = str[i];
str[i] = str[j];
str[j] = temp;
}
}
}
``````
``````public class Solution {
/**
* @param str: An array of char
* @param offset: An integer
* @return: nothing
*/
public void rotateString(char[] str, int offset) {
if(str.length <=0)
return;
offset = offset % str.length;
char[] str1 = new char[str.length + offset];
for(int i = str1.length-1;i>= offset; i--){
str1[i] = str[i - offset];
}

for(int i = 0; i< offset; i++){
str1[i] = str1[i+str.length];
}
for(int i = 0; i< str.length; i++){
str[i] = str1[i];
}
}
}
``````