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+# [74. Search a 2D Matrix](https://leetcode.com/problems/search-a-2d-matrix/)
+# 思路
+就是一个二分查找。将矩阵展开成一个很长的有序数组，下标low从0开始，下标high从`m*n - 1`开始，然后二分查找，中间下标mid对应的元素为`matrix[mid / n][mid % n]`。     
+时间复杂度O(log(mn))，空间复杂度O(1)
+
+# C++
+``` C++
+class Solution {
+public:
+    bool searchMatrix(vector<vector<int>>& matrix, int target) {
+        if(matrix.empty()) return false;
+        int m = matrix.size(), n = matrix[0].size();
+        int low = 0, high = m * n - 1;
+        while(low <= high){
+            int mid = (high - low) / 2 + low;
+            int tmp = matrix[mid / n][mid % n];
+            if(tmp == target) return true;
+            if(tmp < target) low = mid + 1;
+            else high = mid - 1;
+        }
+        return false;
+    }
+};
+```