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.
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.
Solution 1: Two Pointers
Start from first row + last column, if the current value is larger than target, –column; if smaller then ++row.
e.g.
1. r = 0, c = 4, v = 15, 15 > 5 => –c
2. r = 0, c = 3, v = 11, 11 > 5 => –c
3. r = 0, c = 2, v = 7, 7 > 5 => –c
4. r = 0, c = 1, v = 4, 4 < 5 => ++r
5. r = 1, c = 1, v = 5, 5 = 5, found it!
Time complexity: O(m + n)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: bool searchMatrix(vector<vector<int>>& matrix, int target) { if (matrix.empty() || matrix[0].empty()) return false; const int m = matrix.size(); const int n = matrix[0].size(); int r = 0; int c = matrix[0].size() - 1; while (r < m && c >= 0) { if (matrix[r][c] == target) return true; else if (matrix[r][c] > target) --c; else ++r; } return false; } }; |
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