378. Kth Smallest Element in a Sorted Matrix

最新推荐文章于 2020-07-02 20:09:32 发布

wdlsjdl2

最新推荐文章于 2020-07-02 20:09:32 发布

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本文链接：https://blog.youkuaiyun.com/wdlsjdl2/article/details/52083901

本文介绍了一种使用优先队列解决寻找矩阵中第K小元素的方法。该矩阵的每一行和每一列都是按升序排列。通过维护一个优先队列来确保每次取出的都是当前最小的未访问过的元素，并加入其相邻的未访问过的元素。

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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.

堆/优先队列，从最小开始，每次把相邻的加入优先队列，直到K个为止

public class Solution {
    public int kthSmallest(int[][] matrix, int k)
	{
		int m=matrix.length;
		if(m==0)
			return 0;
		
		int n=matrix[0].length;
		int cnt=0;
		boolean[][] used=new boolean[m][n];
		PriorityQueue<kthsNode> heap=new PriorityQueue<>();
		heap.add(new kthsNode(0, 0, matrix[0][0]));
		used[0][0]=true;
		kthsNode node=null;
		int[] xoff={0,1};
		int[] yoff={1,0};
		while(cnt<k)
		{
			node=heap.poll();
			int x=node.x;
			int y=node.y;
			for(int i=0;i<2;i++)
			{
				int xx=x+xoff[i];
				int yy=y+yoff[i];
				if(xx>=0&&xx<m&&yy>=0&&yy<n&&!used[xx][yy])
				{
					used[xx][yy]=true;
					heap.add(new kthsNode(xx, yy, matrix[xx][yy]));
				}
			}
			cnt++;
		}
		
		return node.val;
	}
}


class kthsNode implements Comparable<kthsNode>
{
	int x,y;
	int val;
	public kthsNode(int x,int y,int val)
	{
		this.x=x;
		this.y=y;
		this.val=val;
		// TODO Auto-generated constructor stub
	}
	@Override
	public int compareTo(kthsNode o)
	{
		// TODO Auto-generated method stub
		return val-o.val;
	}
}