kth-smallest-number-in-sorted-matrix
Find the kth smallest number in at row and column sorted matrix.
[
[1 ,5 ,7],
[3 ,7 ,8],
[4 ,8 ,9],
]
注意点：
-
Comparator<Pair>的实现
-
PriorityQueue poll add
1 import java.util.*; 2 3 class Pair{ 4 public int x, y, val; 5 public Pair(int x, int y, int val){ 6 this.x = x; 7 this.y = y; 8 this.val = val; 9 } 10 } 11 12 class PairComparator implements Comparator<Pair> { 13 public int compare(Pair a, Pair b){ 14 return a.val - b.val; 15 } 16 } 17 18 public class Solution { 19 /** 20 * @param matrix: a matrix of integers 21 * @param k: an integer 22 * @return: the kth smallest number in the matrix 23 */ 24 public int kthSmallest(int[][] matrix, int k) { 25 // write your code here 26 int[] dx = {0, 1}; 27 int[] dy = {1, 0}; 28 int m = matrix.length; 29 int n = matrix[0].length; 30 boolean[][] visited = new boolean[m][n]; 31 PriorityQueue<Pair> pq = new PriorityQueue<>(k, new PairComparator()); 32 pq.add(new Pair(0, 0, matrix[0][0])); 33 34 for (int i = 0; i < k - 1; i++){ 35 Pair cur = pq.poll(); 36 for (int j = 0; j < 2; j++){ 37 int x = cur.x + dx[j]; 38 int y = cur.y + dy[j]; 39 if (x < m && y < n && !visited[x][y]){ 40 pq.add(new Pair(x, y, matrix[x][y])); 41 visited[x][y] = true; 42 43 } 44 } 45 } 46 return pq.peek().val; 47 } 48 }