LeetCode 솔루션					분류
				
						[7/24] 240. Search a 2D Matrix II
본문
Medium
7925131Add to ListShareWrite 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
Accepted
633,101
Submissions
1,290,293
관련자료
- 
			링크
			댓글 1
					
			학부유학생님의 댓글
- 익명
- 작성일
					
										
					Runtime: 241 ms, faster than 60.46% of Python3 online submissions for Search a 2D Matrix II.
Memory Usage: 20.5 MB, less than 13.29% of Python3 online submissions for Search a 2D Matrix II.
				
													
								Memory Usage: 20.5 MB, less than 13.29% of Python3 online submissions for Search a 2D Matrix II.
class Solution:
    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
        ROW, COL = len(matrix), len(matrix[0])
        
        r,c = 0, COL-1
        
        while r < ROW and c >= 0:
            if matrix[r][c] < target:
                r += 1
            elif matrix[r][c] > target:
                c -= 1
            else:
                return True
        
        return False 
								 
							








