LeetCode:Rotate Image

题目：

将一个矩阵顺时针旋转90度，要求空间复杂度为O(1)

例子:

Given input matrix = 
[
  [1,2,3],
  [4,5,6],
  [7,8,9]
],

rotate the input matrix in-place such that it becomes:
[
  [7,4,1],
  [8,5,2],
  [9,6,3]
]

思路：

说来惭愧，这个题一开始拿到没思路，因为没有暴力方法，搁置几天后再看发现其实可以通过翻转来解题，先以反对角线为轴旋转，然后在水平翻转即为旋转90度的效果。

[1,2,3]             [9,6,3]            [7,4,1]
[4,5,6]   -->    [8,5,2]    -->        [8,5,2]
[7,8,9]             [7,4,1]            [9,6,3]

代码：

做完居然是标答超过100%

class Solution {
    public void rotate(int[][] matrix) {

        if( matrix == null || matrix.length < 1){
            return;
        }

        int rowLength = matrix.length;
        int colLength = matrix[0].length;
        for(int i = 0; i < rowLength - 1;i++ ){
            for(int j = 0; j < rowLength-i-1; j++){
                swap(matrix,i,j,colLength-j-1,rowLength-i-1);
            }
        }

        for(int i = 0; i < rowLength/2; i++){
            for(int j = 0; j < colLength; j++){
                swap(matrix,i,j,rowLength-i-1,j);
            }
        }
    }

    private void swap(int[][] nums,int fromX,int fromY,int toX,int toY){
        int temp = nums[fromX][fromY];
        nums[fromX][fromY] = nums[toX][toY];
        nums[toX][toY] = temp;
    }
}

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