程序员面试金典 - 面试题 01.07. 旋转矩阵（一次遍历+位运算）

示例 1:
给定 matrix = 
[
  [1,2,3],
  [4,5,6],
  [7,8,9]
],

原地旋转输入矩阵，使其变为:
[
  [7,4,1],
  [8,5,2],
  [9,6,3]
]

示例 2:
给定 matrix =
[
  [ 5, 1, 9,11],
  [ 2, 4, 8,10],
  [13, 3, 6, 7],
  [15,14,12,16]
], 

原地旋转输入矩阵，使其变为:
[
  [15,13, 2, 5],
  [14, 3, 4, 1],
  [12, 6, 8, 9],
  [16, 7,10,11]
]

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
    	int i, j, a, b;
    	for(i = 0; i < matrix.size(); ++i)
    	{	//对角线交换
    		for(j = 0; j < i; ++j)
    			swap(matrix[i][j], matrix[j][i]);
    	}
    	for(i = 0; i < matrix.size(); ++i)
    	{	//每行反转
    		a = 0, b = matrix[0].size()-1;
    		while(a < b)
    			swap(matrix[i][a++], matrix[i][b--]);
    	}
    }
};

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

    inline void swp(int&a, int&b)
    {
    	// b = a^b;
    	// a = a^b;
    	// b = a^b;
        //或者一行代替
        a ^= b ^= a ^= b;
    }
};