378. Kth Smallest Element in a Sorted Matrix

最新推荐文章于 2020-06-29 10:46:52 发布

weixin_33835103

最新推荐文章于 2020-06-29 10:46:52 发布

阅读量43

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/xiuxiu55/p/6508342.html

本文介绍了一种高效算法，用于在一个n x n的矩阵中寻找第k小的元素，该矩阵的每一行和每一列都按升序排列。通过使用优先队列实现堆来维护一个大小为k的最小堆，确保最终堆顶元素即为所求。

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.

class Solution {
public:
    int kthSmallest(vector<vector<int>>& matrix, int k) {
        if (matrix.size() == 0)    return 0;
        priority_queue<int> heap;
    
        int rows = matrix.size();
        int columns = matrix[0].size();
    
        for (int i = 0; i < rows; i++)
            for (int j = 0; j < columns; j++) {
                if (heap.size() < k)   heap.push(matrix[i][j]);
                else {
                    if (heap.top() < matrix[i][j]) break;
                    else {
                        heap.push(matrix[i][j]);
                        heap.pop();
                    }
                }
            }
        return heap.top();
    }
};