题目描述

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.

样例

Example 1:

Input: nums1 = [1,7,11], nums2 = [2,4,6], k = 3
Output: [[1,2],[1,4],[1,6]] 
Explanation: 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:

Input: nums1 = [1,1,2], nums2 = [1,2,3], k = 2
Output: [1,1],[1,1]
Explanation: 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:

Input: nums1 = [1,2], nums2 = [3], k = 3
Output: [1,3],[2,3]
Explanation: All possible pairs are returned from the sequence: [1,3],[2,3]

算法1

(Greedy) $O(n^2)$

在candidates中选出和最小的数对，将其index向右推进一步，重复操作k次，得到最小的k个数对。用重新定义的priority_queue取得最小的数对和，vector[HTML_REMOVED]中存放{nums1[i]，nums2[j],j},每次操作j++将nums2中的数右移一位。

时间复杂度

O(n1+klogk)
space O(k);

参考文献

C++ 代码

struct comp{
        bool operator()(vector<int> &a, vector<int>&b)
        {
            return a[0]+a[1]>b[0]+b[1]||(a[0]+a[1]==b[0]+b[1] && a[2]>b[2]);
        }
    };
class Solution {
public:
    vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
        int n1=nums1.size(), n2=nums2.size();  
        if(n1==0 || n2==0) return {};
        vector<vector<int>> res;
        priority_queue<vector<int>, vector<vector<int>>, comp> q;
        for(int k=0;k<n1;k++)
        {
            q.push({nums1[k], nums2[0], 0});
        }
        while(k-- && !q.empty())
        {
            auto t=q.top(); q.pop();
            res.push_back({t[0],t[1]});
            int idx=t[2];
            if(idx<n2-1)
            {
                idx++;
                q.push({t[0],nums2[idx], idx});
            }
        }
        return res; 
    }
};
/*
nums1 = [1,7,11], 
nums2 = [2,4,6],    k = 3

1,2,0   0->1  1,4,1  1->2  1,6,2
7,2,0
11,2,0
*/