lintcode8 - rotate string - easy

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"

Challenge

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

 

三步反转法。

如果点a是length -1 - offset那个index，反转[0, a]，反转[a+1, end]，反转[0,end]

 

细节：

一开始对offset做一个offset = offset % length的处理，可以避免算index越界。 

 

 

我的实现

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 - 1 - offset);
        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;
        }
    }
}