利用离散数学实现 字符串循环移动 的高效优雅算法

本算法详细说明：http://blog.csdn.net/maoyuanming0806/article/details/78932840

代码：
	/**反转。把字符数组从from位置到to位置的数反转*/
	static void revertStr(char[] str, int from, int to){
		if(to - from < 1){
			return;
		}
		for(int i = from; i < (to + from + 1)/2;i++){
			swapChar(str,i,to -(i - from));
		}
	}
	
	static void swapChar(char[] s, int i, int j) {
		char temp = s[i];
		s[i] = s[j];
		s[j] = temp;
	}
	
	/**字符数组str中moveStrHead位置到moveStrEnd位置的字符向右循环移动moveLength位*/
	static void circleLeftMove(char[] str, int moveStrHead, int moveStrEnd, int moveLength){
		revertStr(str, moveStrHead, moveStrEnd);
		revertStr(str, moveStrEnd+1, moveStrEnd + moveLength);
		revertStr(str, moveStrHead, moveStrEnd + moveLength);
	}



//===========测试：
	public static void main(String[] args) {
		String test = "123456789";
		char[] charArray = test.toCharArray();
		circleLeftMove(charArray, 0, 2, 2);
		System.out.println(charArray);
	}

    输出：451236789