题目描述

题目链接：373. Find K Pairs with Smallest Sums

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.

例子

例子 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]

例子 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]

例子 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 <= 10^4
• -10^9 <= nums1[i], nums2[i] <= 10^9
• nums1 and nums2 both are sorted in ascending order.
• 1 <= k <= 1000

解题思路

暴力解法是遍历获得所有可能的数对，然后进行排序。稍微好一点的方法则是在获取数对的过程中维护一个优先队列，这样可以降低排序的时间复杂度。代码如下：

#include <queue>
#include <vector>

class Solution {
public:
    std::vector<std::vector<int>> kSmallestPairs(std::vector<int>& nums1,
                                                 std::vector<int>& nums2,
                                                 int k) {
        auto comp = [](const std::vector<int>& lhs,
                       const std::vector<int>& rhs) {
            return (lhs[0] + lhs[1]) < (rhs[0] + rhs[1]);
        };

        std::priority_queue<std::vector<int>, std::vector<std::vector<int>>,
                            decltype(comp)>
            pq(comp);

        for (auto n1 : nums1) {
            for (auto n2 : nums2) {
                pq.push({n1, n2});
                if (pq.size() > k) {
                    pq.pop();
                }
            }
        }

        std::vector<std::vector<int>> ret;
        while (!pq.empty()) {
            ret.push_back(pq.top());
            pq.pop();
        }

        return ret;
    }
};

• 时间复杂度: O(n1n2log(k))
• 空间复杂度: O(k)

GitHub 代码同步地址： 373.FindKPairsWithSmallestSums.cpp

其他题目： GitHub: Leetcode-C++-Solution 博客： Leetcode-Solutions