Leetcode 373. Find K Pairs with Smallest Sums

373. Find K Pairs with Smallest Sums

Total Accepted: 1453
Total Submissions: 5789
Difficulty: Medium

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]

思路：

方法一：直接将pair(num1,num2)插入priority_queue中，将优先队列的元素按num1+num2升序排列，最后取前min(nums1.size()*nums2.size(),k)个返回。

方法二：方法一的优化。

方法三：对nums1中的每个元素num1保留对应的nums2元素num2位置的数组index。每次遍历nums1中的每个元素，对于nums1中的元素nums1[i]，在nums2中搭配nums2[index[nums1[i]]]，取最小的nums1[i]+nums2[index[nums1[i]]]，然后将(nums1[i],nums2[index[nums1[i]]])插入到结果的向量中。

关于priority_queue的用法可以参考priority_queue的用法。

代码：

方法一：

 1 class Solution {
 2 public:
 3     struct cmp{
 4         bool operator()(pair<int, int> num1,pair<int, int> num2){
 5             //priority_queue的比较函数用的是less函数：
 6             //返回true时，num1的优先级低于num2（num1排在num2的后面）
 7             return num1.first+num1.second>num2.first+num2.second;
 8         }
 9     };
10     vector<pair<int, int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
11         vector<pair<int, int>> res;
12         priority_queue<pair<int, int>,vector<pair<int, int>>,cmp> pq;
13         int i,j,m=nums1.size(),n=nums2.size();
14         for(i=0;i<m;i++){
15             for(j=0;j<n;j++){
16                 pq.push(make_pair(nums1[i],nums2[j]));
17             }
18         }
19         while(k-->0&&pq.size()>0){
20             res.push_back(pq.top());
21             pq.pop();
22         }
23         return res;
24     }
25 };

方法二：

 1 class Solution {
 2 public:
 3     struct cmp{
 4         bool operator()(pair<int, int>& num1, pair<int, int>& num2){
 5             return num1.first + num1.second < num2.first + num2.second;
 6         }
 7     };
 8     vector<pair<int, int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
 9         vector<pair<int, int>> res;
10         priority_queue<pair<int,int>, vector<pair<int, int> >, cmp> pq;
11         for(int i = 0; i < min((int)nums1.size(), k); i++){
12             for(int j = 0; j < min((int)nums2.size(), k); j++){
13                 if(pq.size() < k)
14                     pq.push(make_pair(nums1[i], nums2[j]));
15                 else if(nums1[i] + nums2[j] < pq.top().first + pq.top().second){
16                     pq.push(make_pair(nums1[i], nums2[j]));
17                     pq.pop();
18                 }
19             }
20         }
21         while(!pq.empty()){
22             res.insert(res.begin(),pq.top());
23             pq.pop();
24         }
25         return res;
26     }
27 };

方法三：

 1 class Solution {
 2 public:
 3 vector<pair<int, int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
 4         vector<pair<int, int>> res;
 5         int i,j,m=nums1.size(),n=nums2.size();
 6         vector<int> index(m,0);
 7         k=min(k,m*n);
 8         while(k-->0){
 9             int temp;
10             long long minv=LONG_MAX;
11             for(i=0;i<m;i++){
12                 if(index[i]<n&&minv>nums1[i]+nums2[index[i]]){
13                     minv=nums1[i]+nums2[index[i]];
14                     temp=i;
15                 }
16             }
17             res.push_back(make_pair(nums1[temp],nums2[index[temp]]));
18             index[temp]++;
19         }
20         return res;
21     }
22 };

Leetcode 373. Find K Pairs with Smallest Sums

373. Find K Pairs with Smallest Sums

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