## [Leetcode]373. Find K Pairs with Smallest Sums(C++)

### 题目描述

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.

### 例子

#### 例子 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]`

#### 例子 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]`

#### 例子 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 <= 10^4`
• `-10^9 <= nums1[i], nums2[i] <= 10^9`
• `nums1` and `nums2` both are sorted in ascending order.
• `1 <= k <= 1000`

### 解题思路

``````#include <queue>
#include <vector>

class Solution {
public:
std::vector<std::vector<int>> kSmallestPairs(std::vector<int>& nums1,
std::vector<int>& nums2,
int k) {
auto comp = [](const std::vector<int>& lhs,
const std::vector<int>& rhs) {
return (lhs[0] + lhs[1]) < (rhs[0] + rhs[1]);
};

std::priority_queue<std::vector<int>, std::vector<std::vector<int>>,
decltype(comp)>
pq(comp);

for (auto n1 : nums1) {
for (auto n2 : nums2) {
pq.push({n1, n2});
if (pq.size() > k) {
pq.pop();
}
}
}

std::vector<std::vector<int>> ret;
while (!pq.empty()) {
ret.push_back(pq.top());
pq.pop();
}

return ret;
}
};
``````
• 时间复杂度: `O(n1n2log(k))`
• 空间复杂度: `O(k)`

GitHub 代码同步地址： 373.FindKPairsWithSmallestSums.cpp

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