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]
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public class Solution { public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) { List<int[]> ret = new ArrayList<>(); if (nums1.length == 0 || nums2.length == 0 || k == 0) { return ret; } //num2Index keeps the index of next number in nums2, that will pair with the value in nums1 int[] num2Index = new int[nums1.length]; int num1start = 0; while (k-- > 0) { //every time we find a min, we will restart from the first number in nums1. int currentMin = Integer.MAX_VALUE; int found = -1; for (int i = num1start; i < nums1.length; ++i) { if (num2Index[i] >= nums2.length) {//nums[i] has been fully paired. num1start = i; // we only need to start to i for the next min search continue; } int nextMin = nums1[i] + nums2[num2Index[i]]; if (nextMin < currentMin) { currentMin = nextMin; found = i; } } if (found == -1) return ret; //cannot find any more pairs, which mean k is larger than total combination counts ret.add(new int[]{nums1[found], nums2[num2Index[found]]}); num2Index[found]++; //set the corresponding index to next one } return ret; } }
373. Find K Pairs with Smallest Sums
原文：http://www.cnblogs.com/neweracoding/p/5698424.html