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题目

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.

Return the k pairs (u1, v1), (u2, v2), …, (uk, vk) with the smallest sums.

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]

Constraints:

1 <= nums1.length, nums2.length <= 105
-109 <= nums1[i], nums2[i] <= 109
nums1 and nums2 both are sorted in ascending order.
1 <= k <= 1000

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代码

class Solution {
public:
    struct cmp {
        bool operator()(pair<int, int>&a, pair<int, int>&b)
        {
            return a.first+a.second < b.first+b.second;
        }
    };
    vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
        vector<vector<int>> res;
        if(nums1.size() < 1 || nums2.size() < 1)
            return res;
        priority_queue<pair<int, int>, vector<pair<int, int>>, cmp> q;
        for(int i = 0; i<nums1.size() && i < k; i++)
        {
            for(int j = 0; j<nums2.size() && j < k; j++)
            {
                if(q.size() < k)
                {
                    q.push(pair<int, int>(nums1[i], nums2[j]));
                }
                else
                {
                    pair<int, int> t = q.top();
                    if(nums1[i] + nums2[j] < t.first + t.second)
                    {
                        q.push(pair<int, int>(nums1[i], nums2[j]));
                        q.pop();
                    }
                }
            }
        }
        while(!q.empty())
        {
            pair<int, int> t = q.top();
            res.push_back({t.first, t.second});
            q.pop();
        }
        return res;
    }
};