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Given a string and an offset, rotate string by offset. (rotate from left to right)
Given "abcdefg".
offset=0 => "abcdefg"
offset=1 => "gabcdef"
offset=2 => "fgabcde"
offset=3 => "efgabcd"
分析：
利用 （A^TB^T）^T = BA
1 public class Solution { 2 /** 3 * @param str: an array of char 4 * @param offset: an integer 5 * @return: nothing 6 */ 7 public void rotateString(char[] str, int offset) { 8 if (str == null || str.length <= 1) return; 9 offset = offset % str.length; 10 if (offset == 0) return; 11 12 int p = str.length - 1 - offset; 13 14 swapFromIToJ(str, 0, p); 15 swapFromIToJ(str, p + 1, str.length - 1); 16 swapFromIToJ(str, 0, str.length - 1); 17 } 18 19 public void swapFromIToJ(char[] str, int i, int j) { 20 while (i < j) { 21 char temp = str[i]; 22 str[i] = str[j]; 23 str[j] = temp; 24 i++; 25 j--; 26 } 27 } 28 }
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原文地址：http://www.cnblogs.com/beiyeqingteng/p/5645642.html
