You are given two integer arrays nums1 and nums2 sorted in ascending order and an integer k.
给你两个数组nums1和nums2,这两个数组都是递增排列的，还给你一个整数k。
Define a pair (u,v) which consists of one element from the first array and one element from the second array.
定义一个pair(u,v)其中一个数是第一个数组中的，另一个数十第二个数组中的。
Find the k pairs (u1,v1),(u2,v2) ...(uk,vk) with the smallest sums.
找出在所有pair中两数和由小到大排列的前k个pair。
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]
1 class Solution { 2 public: 3 vector<pair<int, int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) { 4 vector<pair<int,int>> ret; 5 if(nums1.empty()||nums2.empty()) return ret; 6 int l1=nums1.size(),l2=nums2.size(); 7 vector<int> mm(l1,0); 8 for(int count=0,low=0;count<k&&mm[l1-1]<l2;){ 9 int tmpi=low,tmpms=nums1[low]+nums2[mm[low]]; 10 for(int i=low+1;i<l1;i++){ 11 if(mm[i]==l2)continue; 12 int tmp=nums1[i]+nums2[mm[i]]; 13 if(tmp<tmpms){tmpms=tmp;tmpi=i;} 14 if(mm[i]==0)break; 15 } 16 ret.push_back(pair<int,int>(nums1[tmpi],nums2[mm[tmpi]])); 17 mm[tmpi]++; 18 if(mm[low]==l2)low++; 19 count++; 20 } 21 return ret; 22 } 23 };
上面是AC代码，不太会写解释，还麻烦，有问题可以消息我。