373. Find K Pairs with Smallest Sums

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]

Credits:

Special thanks to @elmirap and @StefanPochmann for
adding this problem and creating all test cases.

数组是分别有序的！还是多路归并问题，对于num1中的每一个元素都设置一个指针，用于指示组合到num2中的哪一个位置。

每次选出和最小的并推进指针直到无数据可取或者满足K个的要求。

public static List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k)
{
int len1=nums1.length;
int len2=nums2.length;
ArrayList<int[]> retlist=new ArrayList<>();
int[] point=new int[len1];
int cnt=0;
int expect=Math.min(len1*len2, k);
while(cnt<expect)
{
int min=Integer.MAX_VALUE;
int minindex=-1;
for(int i=0;i<len1;i++)
if(point[i]<len2)
{
int num=nums1[i]+nums2[point[i]];
if(num<min)
{
min=num;
minindex=i;
}
}
retlist.add(new int[] {nums1[minindex],nums2[point[minindex]]});
point[minindex]++;
cnt++;
}

return retlist;
}