LeetCode Find K Pairs with Smallest Sums

LeetCode 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]

给定两个从小到大排列的数组,要求选出k个数对(u_i,v_i),其中u_i来自第一个数组nums1,v_i来自第二个数组v_i,使得这k个数对的和最小。

要使得k各数对的和最小,那么这k个数对本身肯定是前k个最小的。遇到要选前k个最小或者最大的问题,一般都用优先队列(堆)来做,比如维护一个大小为k的大顶堆,当堆中元素大于k时,弹出堆顶的最大的元素,也就是堆中维护的一直是当前最小的前k个元素。当所有元素都遍历完时,返回堆中所有元素即可。

优先队列采用STL中的priority_queue,需要自定义比较运算符。

完整代码如下:

class Solution {
private:
	struct P {
		int x, y;
		P(int x_, int y_) :x(x_), y(y_) {};
		bool operator<(const P& p) const{
			return (this->x + this->y) < (p.x + p.y);
		}
		bool operator>(const P& p) const {
			return (this->x + this->y) > (p.x + p.y);
		}
	};
public:
	vector<pair<int, int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
		priority_queue<P> pq; // 大顶堆
		for (int i = 0; i < nums1.size(); ++i) {
			for (int j = 0; j < nums2.size(); ++j) {
				P p(nums1[i], nums2[j]);
				pq.push(p);
				if (pq.size() > k)pq.pop(); // 弹出最大元素
			}
		}
		vector<pair<int, int>> ans;
		while (!pq.empty()) { // 保留的都是最小元素
			ans.push_back(pair<int, int>(pq.top().x, pq.top().y));
			pq.pop();
		}
		return ans;
	}
};

本代码提交AC,用时99MS。

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