[leetcode] 373. Find K Pairs with Smallest Sums-优快云博客

本文介绍了一种算法问题，即从两个有序整数数组中找到总和最小的k对数字组合。通过三种不同的方法实现，包括直接排序、使用映射表以及优化迭代方式，对比了它们在效率上的差异。

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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]

解法一：

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++){
                res.push_back({nums1[i],nums2[j]});
            }
        }
        
        sort(res.begin(), res.end(), [](pair<int,int>&a, pair<int,int>&b){return a.first+a.second<b.first+b.second;});
        if((int)res.size()>k) res.erase(res.begin()+k,res.end());
        return res;
        
    }
};

解法二：

class Solution {
public:
    vector<pair<int, int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
        vector<pair<int, int>> res;
        multimap<int, pair<int, int>> m;
        
        for(int i=0; i<min((int)nums1.size(),k); i++){
            for(int j=0; j<min((int)nums2.size(),k); j++){
                m.insert({nums1[i]+nums2[j], {nums1[i],nums2[j]}});
            }
        }
        
        for(auto it = m.begin(); it!=m.end(); it++){
            res.push_back(it->second);
            if(--k<=0) return res;
        }
        
        return res;
    }
};

解法三：

class Solution {
public:
    vector<pair<int, int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
        int len = min(k, int(nums1.size()*nums2.size()));
        vector<int> idx(nums1.size(),0);
        vector<pair<int,int>> res;
        for(int i=0; i<len; i++){
            int t = 0, sum = INT_MAX;
            for(int j=0; j<nums1.size(); j++){
                if(idx[j]<nums2.size() && nums1[j]+nums2[idx[j]]<sum){
                    t = j;
                    sum = nums1[j]+nums2[idx[j]];
                }
            }
            res.push_back({nums1[t],nums2[idx[t]]});
            ++idx[t];
        }
        return res;
    }
};