LeetCode 378. Kth Smallest Element in a Sorted Matrix (Medium)
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题目描述:
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.
题目大意:给出一个矩阵,从左到右,从上到下都是排序好的(从小到大),找出第K小的元素。
思路:这题除了暴力没有思路…找了一下网上的解法,大意就是扫描每一行,建立一个含k个元素的最大堆,如果堆满了,而当前扫描的元素大于堆顶元素,就直接扫描下一行(因为当前元素后面的一定大于当前元素),如果遇到比堆顶小的元素,就删掉堆顶,插入当前元素,最后堆顶元素即是第k小元素。至于最大堆的实现,直接用优先队列。
c++代码:
class Solution {public: int kthSmallest(vector<vector<int>>& matrix, int k) { priority_queue<int> heap; for (int i = 0; i < matrix.size(); i++) { for (int j = 0; j < matrix[i].size(); j++) { if (heap.size() < k) heap.push(matrix[i][j]); else { if (matrix[i][j] > heap.top()) break; else { heap.pop(); heap.push(matrix[i][j]); } } } } return heap.top(); }};
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