373. Find K Pairs with Smallest Sums

题目链接：

greedy: 先把一组x里面和另外一组y最小元素的组合放进heap，然后每次poll出和最小的，同时放进去有可能成为第二小的组合，即当前y元素的下一个和x元素的组合。

public class Solution {    public List
    
      kSmallestPairs(int[] nums1, int[] nums2, int k) {        if(nums1.length == 0 || nums2.length == 0)  return new ArrayList();                // heap: [nums1[i], nums2[j], i, j]        PriorityQueue
     
       minHeap = new PriorityQueue<>(k, (a, b) -> a[0] + a[1] - (b[0] + b[1]));        // add combination of element from one array with the first element of another        for(int i = 0; i < Math.min(nums1.length, k); i++) {            minHeap.offer(new int[] {nums1[i], nums2[0], i, 0});        }        // greedy        List
      
        result = new ArrayList();        while(k-- > 0 && !minHeap.isEmpty()) {            int[] cur = minHeap.poll();            int i = cur[2], j = cur[3];            result.add(new int[] {cur[0], cur[1]});            if(j + 1 < nums2.length) minHeap.offer(new int[] {nums1[i], nums2[j + 1], i, j + 1});        }                return result;    }}