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kth-smallest-element-in-a-sorted-matrix.py 891 Bytes
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Allen Liu 提交于 2018-10-12 01:33 +08:00 . remove sensitive question description
# Time: O(k * log(min(n, m, k))), with n x m matrix
# Space: O(min(n, m, k))
from heapq import heappush, heappop
class Solution(object):
def kthSmallest(self, matrix, k):
"""
:type matrix: List[List[int]]
:type k: int
:rtype: int
"""
kth_smallest = 0
min_heap = []
def push(i, j):
if len(matrix) > len(matrix[0]):
if i < len(matrix[0]) and j < len(matrix):
heappush(min_heap, [matrix[j][i], i, j])
else:
if i < len(matrix) and j < len(matrix[0]):
heappush(min_heap, [matrix[i][j], i, j])
push(0, 0)
while min_heap and k > 0:
kth_smallest, i, j = heappop(min_heap)
push(i, j + 1)
if j == 0:
push(i + 1, 0)
k -= 1
return kth_smallest
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