74. Search a 2D Matrix(二维二分)

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题目链接
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.
Example 1:
Input:
matrix = [
    [1, 3, 5, 7],
    [10, 11, 16, 20],
    [23, 30, 34, 50]
]
target = 3
Output: true

Example 2:
Input:
matrix = [
    [1, 3, 5, 7],
    [10, 11, 16, 20],
    [23, 30, 34, 50]
]
target = 13
Output: false

一个测试样例：

bool searchMatrix(vector<vector<int>>& matrix, int target) {
        if(matrix.empty() || matrix[0].empty()) return false;
        int n = matrix.size(),m = matrix[0].size(),l = 0,r = n * m - 1;
        while(l < r){
            int mid = l + r >> 1;
            if(matrix[mid / m][mid % m] >= target) r = mid;//将一维数组的下标变成二维数组的下标
            else l = mid + 1;
        }
        if(matrix[l / m][l % m] == target) return true;
        else return false;
    }