方法一：从右上角开始走，时间复杂度为 $O(n + m)$

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int row = 0, col = matrix[0].size() - 1;

        while (row < matrix.size() && col >= 0)
        {
            if (matrix[row][col] == target) return true;
            else if (matrix[row][col] < target) row ++ ;
            else col -- ;
        }

        return false;
    }
};

方法二：二分法，时间复杂度为 $O(log(n * m))$

将整个数组看成一个升序序列

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int n = matrix.size(), m = matrix[0].size();
        int l = 0, r = n * m - 1;

        while (l <= r)                              //注意这里必须是小于等于
        {
            int mid = (l + r) >> 1;
            int mid_val = matrix[mid / m][mid % m]; //根据 mid 计算行列坐标
            if (mid_val == target) return true;
            else 
            {
                if (mid_val > target) r = mid - 1;
                else l = mid + 1;
            }
        }

        return false;
    }
};