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74.py
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74.py
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# 74. Search a 2D Matrix
# 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 from left to right.
# The first integer of each row is greater than the last integer of the previous row.
# Example 1:
# Input:
# matrix = [
# [1, 3, 5, 7],
# [10, 11, 16, 20],
# [23, 30, 34, 50]
# ]
# target = 3
# Output: true
# Example 2:
# Input:
# matrix = [
# [1, 3, 5, 7],
# [10, 11, 16, 20],
# [23, 30, 34, 50]
# ]
# target = 13
# Output: false
from util import equal
from typing import List
def search_line(line: List[int], target: int) -> bool:
left = 0
right = len(line) - 1
while (left <= right):
mid = left + ((right - left) >> 1)
if line[mid] == target:
return True
elif line[mid] > target:
right = mid - 1
elif line[mid] < target:
left = mid + 1
return False
class Solution:
def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
if len(matrix) < 1:
return False
if len(matrix) == 1:
return search_line(matrix[0], target)
left = 0
right = len(matrix) - 1
while (left <= right):
mid = left + ((right - left) >> 1)
if matrix[mid][0] == target:
return True
elif matrix[mid][0] > target:
right = mid - 1
elif matrix[mid][0] < target:
left = mid + 1
return search_line(matrix[mid - 1], target) or search_line(
matrix[mid], target)
def main():
matrix = [[1, 3, 5, 7]]
equal(Solution().searchMatrix(matrix, 3), True)
matrix = [[22, 2, 2, 2, 2, 2]]
equal(Solution().searchMatrix(matrix, 3), False)
matrix = [[1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50]]
equal(Solution().searchMatrix(matrix, 16), True)
matrix = [[1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50]]
equal(Solution().searchMatrix(matrix, 3), True)
matrix = [[1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50]]
equal(Solution().searchMatrix(matrix, 13), False)
if __name__ == '__main__':
main()