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 in ascending from left to right.
- Integers in each column are sorted in ascending from top to bottom.
For example,
Consider the following matrix:
[ [1, 4, 7, 11, 15], [2, 5, 8, 12, 19], [3, 6, 9, 16, 22], [10, 13, 14, 17, 24], [18, 21, 23, 26, 30] ]
Given target = 5, return true.
Given target = 20, return false.
Analyse: Consider the top-right element. If it's smaller than target, then the first row is eliminated. If it's larger than the target, then the last column is eliminated.
Runtime: 280ms.
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 row = 0, col = n - 1; 8 9 while(row < m && col >= 0){ 10 if(matrix[row][col] < target) row++; 11 else if(matrix[row][col] > target) col--; 12 else return true; 13 } 14 return false; 15 } 16 };