373. Find K Pairs with Smallest Sums

最新推荐文章于 2019-02-12 19:46:38 发布

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最新推荐文章于 2019-02-12 19:46:38 发布

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本文链接：https://blog.youkuaiyun.com/wdlsjdl2/article/details/51913760

本文介绍了一种算法，用于从两个已排序的整数数组中找到具有最小和的K对组合。通过为第一个数组的每个元素设置指向第二个数组的指针，并逐步移动这些指针来实现。示例展示了如何返回具有最小和的前K对。

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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:

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;
	}