Find K Pairs with Smallest Sums

题目链接：https://leetcode.com/problems/find-k-pairs-with-smallest-sums/

题目：

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:

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:

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:

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]

思路：

刚开始没想用heap做，记录几个下标，维护遍历过程中可能最小的pair，结果在更新下标时，越做越麻烦，遂放弃，参考别人的做法：

用最小堆保存遍历过程中的pairs。

当遍历到(i,j)时，继续往下遍历，相邻的结点是(i+1,j)，(i,j+1)

因为是有序数组，这两个pair是较小的，加入堆中。

算法：

public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
List<int[]> res = new ArrayList<int[]>();
boolean visit[][] = new boolean[nums1.length][nums2.length];
Queue<int[]> heap = new PriorityQueue<int[]>(new Comparator<int[]>(){
public int compare(int[] i, int[] j) {
return (nums1[i[0]] + nums2[i[1]] -( nums1[j[0]] + nums2[j[1]]));
}
});

heap.add(new int[] { 0, 0 });
visit[0][0] = true;

while (!heap.isEmpty() && res.size() < k) {
int d[] = heap.poll();
res.add(new int[] { nums1[d[0]], nums2[d[1]] });

if (d[1] + 1 < nums2.length && visit[d[0]][d[1] + 1] == false) {
heap.add(new int[] { d[0], d[1] + 1 });
visit[d[0]][d[1] + 1] = true;
}
if (d[0] + 1 < nums1.length && visit[d[0]+1][d[1]] == false) {
heap.add(new int[] { d[0]+1, d[1]});
visit[d[0]+1][d[1]] = true;
}
}
return res;
}