LeetCode: Find K Pairs with Smallest Sums

最新推荐文章于 2019-06-24 21:58:06 发布

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最新推荐文章于 2019-06-24 21:58:06 发布

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分类专栏： LeetCode 文章标签： leetcode Find K Pairs with Sm array heap priority_queue

本文链接：https://blog.youkuaiyun.com/u012309913/article/details/52311921

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本文介绍了一种高效算法，用于从两个已排序整数数组中找出所有可能元素对中和最小的k对。通过示例详细解释了算法的实现过程，并提供了一种创新的解决方案，利用索引来追踪组合。

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) ...(u_k,v_k) 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]

题目解析：找出两个有序数组里面元素组合和最小的k对数。

这里，需要注意的是是边界问题。

方法很多，最小堆，优先队列等都可以，这里提供另一种思路：用一个数组存储每个nums1对应元素在nums2中组合索引

源代码：

class Solution {
public:
    vector<pair<int, int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
        vector<pair<int, int> > res;
        int m = nums1.size();
        int n = nums2.size();
        
        int cnt = min(k, m * n);
        vector<int> idx(m, 0);
        
        while(cnt > 0)
        {
            int minSum = INT_MAX;
            int curIdx = 0;
            for (int i = 0; i < m; ++i) {
                if (idx[i] < n && nums1[i] + nums2[idx[i]] <= minSum) {
                    minSum = nums1[i] + nums2[idx[i]];
                    curIdx = i;
                }
            }
            res.push_back(make_pair(nums1[curIdx], nums2[idx[curIdx]]));
            idx[curIdx]++;
            cnt--;
        }
        
        return res;
        
    }
};