leetcode 373. Find K Pairs with Smallest Sums

最新推荐文章于 2021-11-07 10:52:29 发布

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最新推荐文章于 2021-11-07 10:52:29 发布

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本文介绍了一种使用排序映射来解决寻找两个有序整数数组中具有最小和的k对元素的问题的方法。通过示例展示了如何从给定数组中找出k对元素，其和最小，并按顺序返回这些对。

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.

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]

1、疯狂利用map的排序功能。

class Solution {
public:
    vector<pair<int, int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) 
    {
        vector<pair<int, int>> ret;
        multimap<int, pair<int, int>> mp;
        for (auto i : nums1)
        {
            for (auto j : nums2)
            {
                mp.insert(pair<int,pair<int, int> >(i+j, make_pair(i, j)));
            }
        }
        for (auto it = mp.begin(); it != mp.end(); it++)
        {
            ret.push_back(it->second);
            if (--k == 0) break;
        }
        return ret;
    }
};