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373. Find K Pairs with Smallest Sums

时间:2018-11-06 11:22:59      阅读:191      评论:0      收藏:0      [点我收藏+]

标签:ems   tco   and   empty   oss   discus   efi   Plan   asc   

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:
Input: nums1 = [1,7,11], nums2 = [2,4,6], k = 3
Output: [[1,2],[1,4],[1,6]] 
Explanation: 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:
Input: nums1 = [1,1,2], nums2 = [1,2,3], k = 2
Output: [1,1],[1,1]
Explanation: 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:
Input: nums1 = [1,2], nums2 = [3], k = 3
Output: [1,3],[2,3]
Explanation: All possible pairs are returned from the sequence: [1,3],[2,3]






https://leetcode.com/problems/find-k-pairs-with-smallest-sums/discuss/84551/simple-Java-O(KlogK)-solution-with-explanation


class Solution {
    public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
        PriorityQueue<int[]> que = new PriorityQueue<>((a,b)->a[0]+a[1]-b[0]-b[1]);
        List<int[]> res = new ArrayList<>();
        if(nums1.length==0 || nums2.length==0 || k==0) return res;
        for(int i=0; i<nums1.length && i<k; i++) que.offer(new int[]{nums1[i], nums2[0], 0});
        while(k-- > 0 && !que.isEmpty()){
            int[] cur = que.poll();
            res.add(new int[]{cur[0], cur[1]});
            if(cur[2] == nums2.length-1) continue;
            que.offer(new int[]{cur[0],nums2[cur[2]+1], cur[2]+1});
        }
        return res;
    }
}

 

373. Find K Pairs with Smallest Sums

标签:ems   tco   and   empty   oss   discus   efi   Plan   asc   

原文地址:https://www.cnblogs.com/tobeabetterpig/p/9913095.html

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