# 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]```

```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;
}
};
```