# [LeetCode] 373. Find K Pairs with Smallest Sums 找和最小的K对数字

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.

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 <= 104
• -109 <= nums1[i], nums2[i] <= 109
• nums1 and nums2 both are sorted in ascending order.
• 1 <= k <= 1000

Credits:
Special thanks to @elmirap and @StefanPochmann for adding this problem and creating all test cases.

class Solution {
public:
vector<pair<int, int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
vector<pair<int, int>> res;
for (int i = 0; i < min((int)nums1.size(), k); ++i) {
for (int j = 0; j < min((int)nums2.size(), k); ++j) {
res.push_back({nums1[i], nums2[j]});
}
}
sort(res.begin(), res.end(), [](pair<int, int> &a, pair<int, int> &b){return a.first + a.second < b.first + b.second;});
if (res.size() > k) res.erase(res.begin() + k, res.end());
return res;
}
};

class Solution {
public:
vector<pair<int, int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
vector<pair<int, int>> res;
multimap<int, pair<int, int>> m;
for (int i = 0; i < min((int)nums1.size(), k); ++i) {
for (int j = 0; j < min((int)nums2.size(), k); ++j) {
m.insert({nums1[i] + nums2[j], {nums1[i], nums2[j]}});
}
}
for (auto it = m.begin(); it != m.end(); ++it) {
res.push_back(it->second);
if (--k <= 0) return res;
}
return res;
}
};

class Solution {
public:
vector<pair<int, int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
vector<pair<int, int>> res;
priority_queue<pair<int, int>, vector<pair<int, int>>, cmp> q;
for (int i = 0; i < min((int)nums1.size(), k); ++i) {
for (int j = 0; j < min((int)nums2.size(), k); ++j) {
if (q.size() < k) {
q.push({nums1[i], nums2[j]});
} else if (nums1[i] + nums2[j] < q.top().first + q.top().second) {
q.push({nums1[i], nums2[j]}); q.pop();
}
}
}
while (!q.empty()) {
res.push_back(q.top()); q.pop();
}
return res;
}
struct cmp {
bool operator() (pair<int, int> &a, pair<int, int> &b) {
return a.first + a.second < b.first + b.second;
}
};
};

class Solution {
public:
vector<pair<int, int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
vector<pair<int, int>> res;
int size = min(k, int(nums1.size() * nums2.size()));
vector<int> idx(nums1.size(), 0);
for (int t = 0; t < size; ++t) {
int cur = 0, sum = INT_MAX;
for (int i = 0; i < nums1.size(); ++i) {
if (idx[i] < nums2.size() && sum >= nums1[i] + nums2[idx[i]]) {
cur = i;
sum = nums1[i] + nums2[idx[i]];
}
}
res.push_back({nums1[cur], nums2[idx[cur]]});
++idx[cur];
}
return res;
}
};

Github 同步地址：

https://github.com/grandyang/leetcode/issues/373

https://leetcode.com/problems/find-k-pairs-with-smallest-sums/

https://leetcode.com/problems/find-k-pairs-with-smallest-sums/discuss/84655/c-solution

https://leetcode.com/problems/find-k-pairs-with-smallest-sums/discuss/84653/c-idea-of-using-multimap

https://leetcode.com/problems/find-k-pairs-with-smallest-sums/discuss/84551/simple-Java-O(KlogK)-solution-with-explanation

LeetCode All in One 题目讲解汇总(持续更新中...)

posted @ 2016-07-08 13:39  Grandyang  阅读(16552)  评论(0编辑  收藏  举报