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