HackMD
373. 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<= 105 - -109 <=
nums1[i],nums2[i]<= 109 nums1andnums2both are sorted in ascending order.- 1 <=
k<= 104
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;
}
};
