373. Find K Pairs with Smallest Sums

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You are given two integer arrays nums1 and nums2 sorted in ascending order and an integer k.

Define a pair (u,v) which consists of one element from the first array and one element from the second array.

Find the k pairs (u1,v1),(u2,v2) …(uk,vk) with the smallest sums.

Example 1:

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Given nums1 = [1,7,11], nums2 = [2,4,6],  k = 3
Return: [1,2],[1,4],[1,6]
The first 3 pairs are returned from the sequence:
[1,2],[1,4],[1,6],[7,2],[7,4],[11,2],[7,6],[11,4],[11,6]

Example 2:

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Given nums1 = [1,1,2], nums2 = [1,2,3],  k = 2
Return: [1,1],[1,1]
The first 2 pairs are returned from the sequence:
[1,1],[1,1],[1,2],[2,1],[1,2],[2,2],[1,3],[1,3],[2,3]

Example 3:

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Given nums1 = [1,2], nums2 = [3],  k = 3
Return: [1,3],[2,3]
All possible pairs are returned from the sequence:
[1,3],[2,3]

解法1：

discussion中的解法。要用到sorted这个条件可以简化不少程序。
由于两个都是sorted，那么第一个array中任何一个数字和第二个数组搭配时，最小的和一定是和第二个数组的第一个元素的配搭。
这样我们可以先把这个组合加入heap中，每次poll出来的时候就把对应的第二个数组的配搭往上挪一个。

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class Solution {
    public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
        List<int[]> res = new ArrayList<>();
        PriorityQueue<int[]> queue = new PriorityQueue<int[]>((a,b) -> (a[0] + a[1] - b[0] - b[1]));
        if (nums1.length == 0 || nums2.length == 0 || k == 0) {
            return res;
        }
        for (int i = 0; i < nums1.length; i++) {
            queue.offer(new int[]{nums1[i], nums2[0], 0});
        }
        while (k-- > 0 && !queue.isEmpty()) {
            int[] cur = queue.poll();
            res.add(new int[]{cur[0], cur[1]});
            int ptr = cur[2];
            if (ptr == nums2.length - 1) continue;
            queue.offer(new int[]{cur[0], nums2[ptr + 1], ptr + 1});
        }
        return res;
    }
}