Search a 2D Matrix

原題如下：
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted from left to right.
The first integer of each row is greater than the last integer of the previous row.
For example,

Consider the following matrix:

[
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
Given target = 3, return true.

這個題其實還是很簡單的，直接暴力也能AC， C++代碼如下：

    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        if(matrix.size() < 1 || matrix[0].size() < 1)
             return false;
        int m = matrix.size(), n = matrix[0].size();
        for(int i=0; i<m; i++){
            for(int j=0; j<n; j++){
                if(matrix[i][j] == target){
                    return true;
                }
            }
        }

        return false;
    }

不過既然有規律可循，那肯定要找到一種高效的方法了，可以考慮把矩陣“展平”，這不就是二分查找嘛。這樣一下子就可以把時間從O(MN)降到了O(Log(MN))了。AC過的C++代碼如下：

    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        if(matrix.size() < 1 || matrix[0].size() < 1)
             return false;
        int m = matrix.size(), n = matrix[0].size();
        int left = 0, right = m * n - 1;

        while(left <= right){
            int mid = (left + right) / 2;
            int i = mid /n, j = mid % n;
            if(matrix[i][j] == target)
               return true;
            else if(matrix[i][j] > target){
                right = mid - 1;
            }else{
                left = mid + 1;
            }
        }

        return false;
    }