373. Find K Pairs with Smallest Sums

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) ...(u_k,v_k) 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]

 class Solution {
 public:
     vector<pair<int, int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
         vector<pair<int,int>> ret;
         if(nums1.empty()||nums2.empty()) return ret;
         int l1=nums1.size(),l2=nums2.size();
         vector<int> mm(l1,);
         for(int count=,low=;count<k&&mm[l1-]<l2;){
             int tmpi=low,tmpms=nums1[low]+nums2[mm[low]];
             for(int i=low+;i<l1;i++){
                 if(mm[i]==l2)continue;
                 int tmp=nums1[i]+nums2[mm[i]];
                 if(tmp<tmpms){tmpms=tmp;tmpi=i;}
                 if(mm[i]==)break;
             }
             ret.push_back(pair<int,int>(nums1[tmpi],nums2[mm[tmpi]]));
             mm[tmpi]++;
             if(mm[low]==l2)low++;
             count++;
         }
         return ret;
     }
 };

上面是AC代码，不太会写解释，还麻烦，有问题可以消息我。