48 Rotate Image

You are given an n x n 2D matrix representing an image.

Rotate the image by 90 degrees (clockwise).

Follow up:
Could you do this in-place?

这 道题虽然操作起来有一点繁琐，但是思路比较简单，就是考察一下数组的基本操作。基本思路是把图片分为行数/2层，然后一层层进行旋转，每一层有上下左右四 个列，然后目标就是把上列放到右列，右列放到下列，下列放到左列，左列放回上列，中间保存一个临时变量即可。因为每个元素搬运的次数不会超过一次，时间复 杂度是O(矩阵的大小)，空间复杂度是O(1)。代码如下： 

public void rotate(int[][] matrix) {
    if(matrix == null || matrix.length==0 || matrix[0].length==0)
        return;
    int layerNum = matrix.length/2;
    for(int layer=0;layer<layerNum;layer++)
    {
        for(int i=layer;i<matrix.length-layer-1;i++)
        {
            int temp = matrix[layer][i];
            matrix[layer][i] = matrix[matrix.length-1-i][layer];
            matrix[matrix.length-1-i][layer] = matrix[matrix.length-1-layer][matrix.length-1-i];
            matrix[matrix.length-1-layer][matrix.length-1-i] = matrix[i][matrix.length-1-layer];
            matrix[i][matrix.length-1-layer] = temp;
        }
    }
}

这种题目就是思路比较简单，不过实现的时候要细心，容易出错。如果面试遇到了还得谨慎对待，尽量不要出错哈。

 

class Solution {
public:
    void rotate(vector<vector<int> > &matrix) {
        const int n = matrix.size();
        
        for (int k = 0; k < n / 2; k++) {
            for (int i = k; i < n - k - 1; i++) {
                int tmp = matrix[k][i];
                matrix[k][i] = matrix[n-1-i][k];
                matrix[n-1-i][k] = matrix[n-1-k][n-1-i];
                matrix[n-1-k][n-1-i] = matrix[i][n-1-k];
                matrix[i][n-1-k] = tmp;
            }
        }        
    }
};