0240. Search a 2D Matrix II
https://leetcode.com/problems/search-a-2d-matrix-ii
Description
Write an efficient algorithm that searches for a target
value in an m x n
integer matrix
. The 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 1:

**Input:** 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]], target = 5
**Output:** true
Example 2:

**Input:** 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]], target = 20
**Output:** false
Constraints:
m == matrix.length
n == matrix[i].length
1 <= n, m <= 300
-109 <= matrix[i][j] <= 109
All the integers in each row are sorted in ascending order.
All the integers in each column are sorted in ascending order.
-109 <= target <= 109
ac
// it's not binary search. strange... binary search would be even more ugly.
// information from description must be useful somehow, try to utilize it.
class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
int row = matrix.length;
// edge case
if (row == 0) return false;
int col = matrix[0].length;
if (col == 0 || matrix[0][0] > target || matrix[row-1][col-1] < target) return false;
// loop
int r = 0, c = col - 1;
while (r < row && c >= 0) {
if (matrix[r][c] == target) {
return true;
} else if (matrix[r][c] < target) {
r++;
} else {
c--;
}
}
// can't find
return false;
}
}
// first row, last col. while row < m && col >= 0
// binRow -> binCol
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