Skip to content

74. Search a 2D Matrix 👍

  • Time: $O(mn\log mn)$
  • Space: $O(1)$
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
class Solution {
 public:
  bool searchMatrix(vector<vector<int>>& matrix, int target) {
    if (matrix.empty())
      return false;

    const int m = matrix.size();
    const int n = matrix[0].size();
    int l = 0;
    int r = m * n;

    while (l < r) {
      const int mid = (l + r) / 2;
      const int i = mid / n;
      const int j = mid % n;
      if (matrix[i][j] == target)
        return true;
      if (matrix[i][j] < target)
        l = mid + 1;
      else
        r = mid;
    }

    return false;
  }
};
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
class Solution {
  public boolean searchMatrix(int[][] matrix, int target) {
    if (matrix.length == 0)
      return false;

    final int m = matrix.length;
    final int n = matrix[0].length;
    int l = 0;
    int r = m * n;

    while (l < r) {
      final int mid = (l + r) / 2;
      final int i = mid / n;
      final int j = mid % n;
      if (matrix[i][j] == target)
        return true;
      if (matrix[i][j] < target)
        l = mid + 1;
      else
        r = mid;
    }

    return false;
  }
}
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
class Solution:
  def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
    if not matrix:
      return False

    m = len(matrix)
    n = len(matrix[0])
    l = 0
    r = m * n

    while l < r:
      mid = (l + r) // 2
      i = mid // n
      j = mid % n
      if matrix[i][j] == target:
        return True
      if matrix[i][j] < target:
        l = mid + 1
      else:
        r = mid

    return False
Back to top