【Leetcode】378. 有序矩阵的第k个元素（Kth Smallest Element in a Sorted Matrix）

Leetcode - 378 Kth Smallest Element in a Sorted Matrix (Medium)

题目描述：矩阵的每一行与每一列均递增，找出第 k 个元素。

matrix = [
   [ 1,  5,  9],
   [10, 11, 13],
   [12, 13, 15]
],
k = 8,

return 13.

解法一：二分查找。

public int kthSmallest(int[][] matrix, int k) {
    int m = matrix.length, n = matrix[0].length;
    int lo = matrix[0][0], hi = matrix[m - 1][n - 1] + 1;
    while (lo < hi) {
        int mid = lo + (hi - lo) / 2;
        int count = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n && matrix[i][j] <= mid; j++) {
                count++;
            }
        }
        if (count < k) lo = mid + 1;
        else hi = mid;
    }
    return lo;
}

解法二：堆。

public class Solution {
    public int kthSmallest(int[][] matrix, int k) {
        int n = matrix.length;
        PriorityQueue<Tuple> pq = new PriorityQueue<Tuple>();
        for(int j = 0; j <= n - 1; j++) 
            pq.offer(new Tuple(0, j, matrix[0][j]));
        for(int i = 0; i < k - 1; i++) {
            Tuple t = pq.poll();
            if(t.x == n - 1) continue;
            pq.offer(new Tuple(t.x + 1, t.y, matrix[t.x + 1][t.y]));
        }
        return pq.poll().val;
    }
}

class Tuple implements Comparable<Tuple> {
    int x, y, val;
    public Tuple (int x, int y, int val) {
        this.x = x;
        this.y = y;
        this.val = val;
    }
    
    @Override
    public int compareTo (Tuple that) {
        return this.val - that.val;
    }
}