Search a 2D Matrix II Write an efficient algorithm that searches for a value target
in an m x n
integer matrix 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 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
Search a 2D Matrix II Solutions
✅Time: O(m+n)
✅Space: O(1)
C++
class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int r = 0;
int c = matrix[0].size() - 1;
while (r < matrix.size() && c >= 0) {
if (matrix[r][c] == target)
return true;
if (matrix[r][c] > target)
--c;
else
++r;
}
return false;
}
};
Java
class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
int r = 0;
int c = matrix[0].length - 1;
while (r <= matrix.length && c >= 0) {
if (matrix[r][c] == target)
return true;
if (matrix[r][c] > target)
--c;
else
++r;
}
return false;
}
}
Python
class Solution:
def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
r = 0
c = len(matrix[0]) - 1
while r < len(matrix) and c >= 0:
if matrix[r][c] == target:
return True
if target < matrix[r][c]:
c -= 1
else:
r += 1
return False
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