Given a matrix of size N x M. For each row the elements are sorted in ascending order, and for each column the elements are also sorted in ascending order. Find the Kth smallest number in it.
Assumptions
the matrix is not null, N > 0 and M > 0
K > 0 and K <= N * M
Examples
{ {1, 3, 5, 7},
{2, 4, 8, 9},
{3, 5, 11, 15},
{6, 8, 13, 18} }
the 5th smallest number is 4
the 8th smallest number is 6
Assumptions
the matrix is not null, N > 0 and M > 0
K > 0 and K <= N * M
1 class Solution {
2 public int kthSmallest(int[][] matrix, int k) {
3 int rows = matrix.length ;
4 int cols = matrix[0].length ;
5 Queue<Cell> minHeap = new PriorityQueue<Cell>(k, new CellComparator());
6 //marker for visited: same rol and col
7 boolean[][] visited = new boolean[rows][cols] ;
8 //expand
9 minHeap.offer(new Cell(0,0,matrix[0][0])) ;
10 visited[0][0] = true ;
11 /*pop k-1 then whats left is the kth smallest:
12 因为行，列 都是单调递增，所以下一个最小的值，肯定在当前点的 左边，或者右边。
13 你不需要考虑加 哪个， pq 又可以帮你做到抛最小的，
14 所以要严格利用这个性质 做 kth 最小，最大的题。
15 */
16 for (int i = 0; i < k-1 ; i++ ) {
17 Cell curr = minHeap.poll();
18 //when dealing with index +1 for array, always checks boundary
19 if (curr.row+1<rows && !visited[curr.row+1][curr.col]) {
20 minHeap.offer(new Cell(curr.row+1, curr.col, matrix[curr.row+1][curr.col]));
21 visited[curr.row+1][curr.col] = true ;
22 }
23 if (curr.col+1<cols && !visited[curr.row][curr.col+1]) {
24 minHeap.offer(new Cell(curr.row, curr.col+1, matrix[curr.row][curr.col+1]));
25 visited[curr.row][curr.col+1] = true ;
26 }
27 }
28 //已经抛出去 k-1 那剩下的 peek 一下就是 第k 个最小的
29 return minHeap.peek().val;
30 }
31 }
32 // PriorityminHeap for custom class requires implements Comparator<T>
33 class Cell{
34 public int row ;
35 public int col ;
36 public int val;
37 public Cell (int row, int col, int val){
38 this.row = row ;
39 this.col = col ;
40 this.val = val ;
41 }
42
43 }
44 class CellComparator implements Comparator<Cell>{
45 @Override
46 public int compare(Cell c1, Cell c2){
47 return c1.val - c2.val ;
48 }
49 }