leetcode 240は、2次元マトリクスII python(1行のコード、複数の解法)を検索する
1828 ワード
すべてのLeetcodeのテーマは不定期にGithubにまとめられ、批判と指摘を歓迎し、交流を討論します.
すべてのLeetcodeのテーマは不定期にGithubにまとめられ、批判と指摘を歓迎し、交流を討論します.
'''
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.
Given target = 20, return false
'''
class Solution:
def searchMatrix(self, matrix, target):
"""
:type matrix: List[List[int]]
:type target: int
:rtype: bool
"""
# Approach one
# if matrix == [] or matrix == [[]]: return False
# length = len(matrix[0]) - 1
# for i in range(len(matrix)):
# if matrix[i][-1] < target: continue
# elif matrix[i][-1] == target: return True
# else:
# for j in range(length):
# if matrix[i][0] > target: return False
# if matrix[i][j] == target: return True
# return False
# Approach two
# if matrix == [] or matrix == [[]]: return False
# row , col = len(matrix) - 1 , 0
# cols = len(matrix[0])
# while row >= 0 and col < cols:
# if matrix[row][col] == target: return True
# elif matrix[row][col] > target:
# row -= 1
# else:
# col += 1
# return False
# Approach three
return any(target in row for row in matrix)
すべてのLeetcodeのテーマは不定期にGithubにまとめられ、批判と指摘を歓迎し、交流を討論します.