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.
For example,
Consider the following matrix:
[ [1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50] ]
Given target = 3
, return true
.
Analyse: Binary Search!!!
Runtime: 16ms.
1 class Solution { 2 public: 3 bool searchMatrix(vector<vector<int>>& matrix, int target) { 4 if(matrix.size() == 0) return false; 5 6 int m = matrix.size(), n = matrix[0].size(); 7 int low = 0, high = m * n - 1; 8 9 while(low <= high){ 10 int mid = (low + high) / 2; 11 int val = matrix[mid / n][mid % n]; 12 13 if(val < target) low = mid + 1; 14 else if(val > target) high = mid - 1; 15 else return true; 16 } 17 return false; 18 } 19 };