Leetcode and Lintcode

Rotate String

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

Example
Given “abcdefg”.

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

Solution:

For the example offset=3 => “efgabcd”.
First reverse “abcd” in “abcdefg”, we can get “dcbaefg”.
Then reverse “efg” in “dcbaefg”, we can get “dcbagfe”.
Finally, reverse the whole string “dcbagfe”, we can get “efgabcd”.

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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; //Important row
        
        reverse(str, 0, str.length - 1 - offset);
        reverse(str, str.length - offset, str.length - 1);
        reverse(str, 0, str.length - 1);
    }
    
    public 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;
        }
    }
}