378. Kth Smallest Element in a Sorted Matrix

和merge k sorted linked list是一样的，不同就是自己要打一个包建一个类，然后建一个pq

 1     class Node{
 2         int x;
 3         int y;
 4         int val;
 5         public Node(int x, int y, int val) {
 6             this.x = x;
 7             this.y = y;
 8             this.val = val;
 9         }
10     }
11     
12     public int kthSmallest(int[][] matrix, int k) {
13         if(matrix.length == 0 || matrix[0].length == 0) {
14             return 0;
15         }
16         int row = matrix.length;
17         int col = matrix[0].length;
18         PriorityQueue<Node> pq = new PriorityQueue<>(new Comparator<Node>() {
19             public int compare(Node n1, Node n2) {
20                 return n1.val - n2.val;
21             }
22         });
23         for(int i = 0; i < matrix.length; i++) {
24             pq.offer(new Node(i, 0,  matrix[i][0]));
25         }
26         while(k > 1) {
27             Node node = pq.poll();
28             k--;
29             if(node.y < row - 1) {
30                 pq.offer(new Node(node.x, node.y + 1, matrix[node.x][node.y + 1]));
31             }
32         }
33         return pq.poll().val;
34     }

 时间复杂度是O(klogn). n = matrix.length

posted @ 2016-10-24 08:31  warmland  阅读(124)  评论(0)  编辑  收藏  举报