方法一：二分法按行遍历查找：

class Solution {
public:bool searchMatrix(vector<vector<int>>& matrix, int target) {for (const auto& row: matrix) {auto it = lower_bound(row.begin(), row.end(), target);if (it != row.end() && *it == target) {return true;}}return false;}
};

方法二：二叉树思想，从左下角元素开始，大于搜索值则行--，小于搜索值则列++，直到等于或者行列达到边界。

class Solution {
public:bool searchMatrix(vector<vector<int>>& matrix, int target) {for (const auto& row: matrix) {auto it = lower_bound(row.begin(), row.end(), target);if (it != row.end() && *it == target) {return true;}}return false;}
};