卖萌的弱渣

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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:

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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]

Solution

  • 运用min-heap 关键是不知道nums1[i+1]+nums2[0] 与 nums1[i]+nums2[j] 之间的关系
  • Java
(Find-K-Pairs-with-Smallest-Sums.java) download
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public class Solution {
    public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
        // min-heap<nums1[i],nums2[j], j> 
        PriorityQueue<int[]> queue = new PriorityQueue<>((a,b)->a[0]+a[1]-b[0]-b[1]);
        List<int[]> ret = new ArrayList<>();
        if(nums1.length==0 || nums2.length==0 || k==0) return ret;
        // 把<nums1[0],nums2[0]>, <nums1[1], nums2[0]> ... <nums1[nums1.length-1], nums[2]>插入到min-heap
        for(int i=0;i<nums1.length && i<k; i++)
            queue.offer(new int[]{nums1[i], nums2[0], 0});

        while(k-- > 0 && !queue.isEmpty()){
            int[] cur = queue.poll();
            ret.add(new int[]{cur[0], cur[1]});
            if(cur[2] == nums2.length-1) continue;
            // 开始<nums1[i],nums[j],j>...<nums1[i],nums[j+1],j+1>
            queue.offer(new int[]{cur[0], nums2[cur[2]+1], cur[2]+1});
        }

        return ret;
    }
}
  • Python
(Find-K-Pairs-with-Smallest-Sums.py) download
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class Solution(object):
    def kSmallestPairs(self, nums1, nums2, k):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :type k: int
        :rtype: List[List[int]]
        """

        # hq: heap_queue<sum,index1,index2>
        hq = []

        def push(index1,index2):
            if index1<len(nums1) and index2<len(nums2):
                heapq.heappush(hq, [nums1[index1]+nums2[index2], index1, index2])

        push(0,0)
        ret = []
        while hq and len(ret) < k:
            # 当前最小的组合
            sums,i,j=heapq.heappop(hq)
            ret.append([nums1[i],nums2[j]])

            # 把下一个nums1[i],nums2[j+1]放进去
            push(i,j+1)

            # 若j==0, 并不知道nums1[i],nums2[0]和nums1[i],nums2[j+1]的关系
            if j==0:
                push(i+1,0)
        return ret