题目描述
题目链接:373. Find K Pairs with Smallest Sums
You are given two integer arrays
nums1
andnums2
sorted in ascending order and an integerk
.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
andnums2
both are sorted in ascending order.1 <= k <= 1000
解题思路
暴力解法是遍历获得所有可能的数对,然后进行排序。稍微好一点的方法则是在获取数对的过程中维护一个优先队列,这样可以降低排序的时间复杂度。代码如下:
#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
其他题目: GitHub: Leetcode-C++-Solution 博客: Leetcode-Solutions