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) …(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]

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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： [cpp] 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; } }; [/cpp] 本代码提交AC，用时99MS。]]>