LeetCode 算法刷题(373)

最新推荐文章于 2023-10-23 13:13:49 发布

wenanshi

最新推荐文章于 2023-10-23 13:13:49 发布

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本文介绍了一个算法问题：给定两个已排序整数数组，找出所有可能配对中和最小的k对。通过示例展示了如何使用优先队列解决此问题，并提供了一段Java代码实现。

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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]

public static void main(String[] args) {
		Test373 test = new Test373();
		int[] nums1 = {1,7,11};
		int[] nums2 = {2,4,6};
		test.kSmallestPairs2(nums1,nums2, 3);

	}
	public List<int[]> kSmallestPairs2(int[] nums1, int[] nums2, int k) {
		List<int[]> res = new LinkedList<>();
        Queue<int[]> queue = new PriorityQueue<>(k,new Comparator<int[]>(){
            public int compare(int[] o1,int[] o2){
                int tmp1 = o1[0]+o1[1];
                int tmp2 = o2[0]+o2[1];

                return tmp1 - tmp2;
            }            
        });

       for(int i = 0;i<nums1.length;i++){
           for(int j = 0;j<nums2.length;j++){
               queue.add(new int[]{nums1[i],nums2[j]});
           }
       }

       while(k-->0){
           int[] tmp = queue.poll();
           if(tmp == null)
                break;
            res.add(tmp);
       }

       return res;
	}