[373. Find K Pairs with Smallest Sums](https://leetcode.com/problems/find-k-pairs-with-smallest-sums/)

### 題目描述

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.

Return *the* `k` *pairs* `(u1, v1), (u2, v2), ..., (uk, vk)` *with the smallest sums.*

### 範例

**Example 1:**

Input: nums1 = [1,7,11], nums2 = [2,4,6], k = 3
Output: [[1,2],[1,4],[1,6]]
Explanation: 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:**

Input: nums1 = [1,1,2], nums2 = [1,2,3], k = 2
Output: [[1,1],[1,1]]
Explanation: 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:**

Input: nums1 = [1,2], nums2 = [3], k = 3
Output: [[1,3],[2,3]]
Explanation: All possible pairs are returned from the sequence: [1,3],[2,3]

**Constraints**:

* 1 <= `nums1.length`, `nums2.length` <= 10^5^
* -10^9^ <= `nums1[i]`, `nums2[i]` <= 10^9^
* `nums1` and `nums2` both are sorted in **ascending order**.
* 1 <= `k` <= 10^4^

### 解答

#### C++

class Solution {
public:
    vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
        ios_base::sync_with_stdio(0); cin.tie(0);
        vector<vector<int>> ans;
        priority_queue<pair<int,int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
        for (int i = 0; i < nums1.size(); i ++) {
            pq.push({nums1[i] + nums2[0], 0});
        }
        while (k -- and not pq.empty()) {
            // cout << "k = " << k << endl;
            // cout << "pq.size() = " << pq.size() << endl;
            const auto [sum, idx2] = pq.top();
            ans.push_back({sum - nums2[idx2], nums2[idx2]});
            pq.pop();
            if (idx2 + 1 < nums2.size()) {
                pq.push({sum - nums2[idx2] + nums2[idx2 + 1], idx2 + 1});
            }
        }
        return ans;
    }
};

### Reference

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