Given a n x n matrix where each of the rows and columns are sorted in ascending order, find the kth smallest element in the matrix.
Note that it is the kth smallest element in the sorted order, not the kth distinct element.
Example:
matrix = [ [ 1, 5, 9], [10, 11, 13], [12, 13, 15] ], k = 8, return 13.
Note:
You may assume k is always valid, 1 ≤ k ≤ n2.
#! /usr/bin/evn python #coding:utf-8 from heapq import heappush, heappop class MinQ(object): def __init__(self): self.h = [] def push(self, value): heappush(self.h, value) def pop(self): return heappop(self.h) class Solution(object): def kthSmallest(self, matrix, k): """ :type matrix: List[List[int]] :type k: int :rtype: int """ assert matrix rows_cnt = len(matrix) min_q = MinQ() for j,c in enumerate(matrix[0]): min_q.push([c, 0, j]) n = 1 val = None while n <= k: val = min_q.pop() i,j = val[1]+1, val[2] if i < rows_cnt: min_q.push([matrix[i][j], i, j]) n += 1 return val[0]
骚年，牢记最小堆的实现！
leetcode 378. Kth Smallest Element in a Sorted Matrix
原文：http://www.cnblogs.com/bonelee/p/6143089.html