373. Find K Pairs with Smallest Sums(unsolved)

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]

解答：

这道题其实就是要学pair怎么使用还有sort函数如何按照自己的想法去写。

就用暴力法解出来就行，取nums1,nums2分别取前k个数或者是他们分别的个数，然后构建pair对，按照pair对的和来排序，从小到大，然后取前k个。

class Solution {
public:
vector<pair<int, int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
vector<pair<int, int>> res;
for(int i=0;i<min((int)nums1.size(),k);i++)
{
for(int j=0;j<min((int)nums2.size(),k);j++)
{
pair<int,int>ret(nums1[i],nums2[j]);
res.push_back(ret);

}
}
sort(res.begin(),res.end(),[](pair<int, int> &a,pair<int, int> &b){ return a.first+a.second<b.first+b.second;});
if(res.size()>k)
res.erase(res.begin()+k,res.end());
return res;

}
};

二刷时自己做出来的方法，时间效率比较慢

class Solution {
public:
vector<pair<int, int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
vector<pair<int, int>> v;
if(nums1.empty()||nums2.empty()) return v;
for(auto n1:nums1){
for(auto n2:nums2){
v.push_back(make_pair(n1,n2));
}
}
sort(v.begin(),v.end(),comp);
vector<pair<int, int>> res;
for(int i=0;(i<k)&&(i<v.size());i++){
res.push_back(v[i]);
}
return res;

}
static bool comp(pair<int, int> a,pair<int, int> b){
return (a.first+a.second) < (b.first+b.second);
}
};