373. Find K Pairs with Smallest Sums (java,优先队列)

题目:

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.

分析:求前k小的组合,组合之间的大小依据是数据对和的大小。同样采用优先队列,队首元素为较小值。

代码:

public class Solution {
    public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
        if(nums1 == null || nums2 == null || k <= 0)
           return null;
        if(nums1.length == 0 || nums2.length == 0)
           return new ArrayList<int[]>();
        if(nums1.length * nums2.length < k)
             k = nums1.length * nums2.length;
        //the priorityqueue 优先队列
        PriorityQueue<Pair> queue = new PriorityQueue<Pair>();
        //result
        List<int[]> list = new ArrayList<int[]>();
        //首先将nums1同nums2[0]的所有组合入队列
        for(int i = 0; i < nums1.length; ++i)
           queue.add(new Pair(i,0,nums1[i],nums2[0]));
        //获取k个最小值
        for(int n = 0; n < k; ++n){
            Pair tmp = queue.poll();
            int[] ele = tmp.pair; //数值对
            list.add(ele); //加入结果集
            if(tmp.j == nums2.length - 1)
               continue;
            queue.add(new Pair(tmp.i,tmp.j + 1,nums1[tmp.i],nums2[tmp.j + 1]));
        }
        return list;
    }
}
//定义pair的数据结构
class Pair implements Comparable<Pair>{
    int[] pair = new int[2]; //定义数组对
    int i; //第一个元素在nums1中的下标
    int j; //第二个元素在nums2中的下标
    public Pair(int i,int j,int e1,int e2){
        pair[0] = e1;
        pair[1] = e2;
        this.i = i;
        this.j = j;
    }

    @Override
    public int compareTo(Pair that){ //升序排列
        return (pair[0] + pair[1]) - (that.pair[0] + that.pair[1]);
    }
}
时间: 2024-10-12 20:40:04

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