Find the kth smallest number in at row and column sorted matrix.
Example
Given k = 4 and a matrix:
[
[1 ,5 ,7],
[3 ,7 ,8],
[4 ,8 ,9],
]
return 5
Challenge
KLog(Min(M,N,K))时间复杂度 K是因为要Poll K次并且同时insert K次，Min(M,N,K)是堆的size，insert的时间是Log(MIN(M,N,K))
思路就是维护一个最小堆，先把第一行，或第一列（本题做法是第一列，之后默认第一列）加入堆中，poll K次，每poll一次之后要把poll的元素的下一个元素加入堆中，本题就是poll的元素的下一列元素。最后一次poll的元素即为所求。因为需要知道每个poll的元素的位置，所以写了个Point Class
public class Solution {
/**
* @param matrix: a matrix of integers
* @param k: an integer
* @return: the kth smallest number in the matrix
*/
// write your code here
public class Point {
public int x, y, val;
public Point(int x, int y, int val) {
this.x = x;
this.y = y;
this.val = val;
}
}
Comparator<Point> comp = new Comparator<Point>() {
public int compare(Point left, Point right) {
return left.val - right.val;
}
};
public int kthSmallest(int[][] matrix, int k) {
if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
return 0;
}
if (k > matrix.length * matrix[0].length) {
return 0;
}
return horizontal(matrix, k);
}
private int horizontal(int[][] matrix, int k) {
Queue<Point> heap = new PriorityQueue<Point>(k, comp);
for (int i = 0; i < Math.min(matrix.length, k); i++) {
heap.offer(new Point(i, 0, matrix[i][0]));
}
for (int i = 0; i < k - 1; i++) {
Point curr = heap.poll();
if (curr.y + 1 < matrix[0].length) {
heap.offer(new Point(curr.x, curr.y + 1, matrix[curr.x][curr.y + 1]));
}
}
return heap.peek().val;
}
}