1.把矩阵当成数组，进行二分查找【行优先】

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        if (matrix.empty()) return false;
        int n = matrix.size(), m = matrix[0].size(), l = 0, r = m * n;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (matrix[mid / m][mid%m] > target) r = mid;
            else l = mid + 1;
        }
        --r;
        return r >= 0 && matrix[r / m][r%m] == target;
    }
};

2.先对行进行二分查找，再对列进行二分查找

class Solution {
public:
    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 = -1, r = n - 1;
        while (l < r) {
            int mid = (l + r + 1) >> 1;
            if (matrix[mid][0] < target) l = mid;
            else r = mid - 1;
        }
        int k = l;
        if (k + 1<n && matrix[k + 1][0] == target) return true;
        if (k < 0) return false;
        l = 0, r = m;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (matrix[k][mid] > target) r = mid;
            else l = mid + 1;
        }
        --r;
        return matrix[k][r] == target;
    }
};