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 = 3Return: [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 = 2Return: [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
     
      
       > kSmallestPairs(vector
       
        & nums1, vector
        
         & nums2, int k) { 4         vector
         
          
           > ret; 5 if(nums1.empty()||nums2.empty()) return ret; 6 int l1=nums1.size(),l2=nums2.size(); 7 vector
           
             mm(l1,0); 8 for(int count=0,low=0;count
            
             
              (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。