Hello everyone! Today's LeetCode Daily problem isFind K Pairs with Smallest Sums. Problem description This problem is commonly solved with with priority queue. Here's a C++ solution for reference. Solution with PQ Many users — and me in particular — initially tried to solve this problem wit...
1classSolution {2publicList<List<Integer>> kSmallestPairs(int[] nums1,int[] nums2,intk) {3//corner case4List<List<Integer>> res =newArrayList<>();5if(nums1 ==null|| nums2 ==null|| nums1.length == 0 || nums2.length == 0) {6returnres;7}89//normal case10PriorityQueue<int[]...
https://github.com/grandyang/leetcode/issues/373 参考资料: https://leetcode.com/problems/find-k-pairs-with-smallest-sums/ https://leetcode.com/problems/find-k-pairs-with-smallest-sums/solutions/4058698/c-priority-queue-k-max-heap-method-most-efficient/ LeetCode All in One 题目讲解汇总(持续...
通过次数 29.6K 提交次数 36.6K 通过率 80.9% 相关标签 贪心数组矩阵 相关企业 提示1 Find the smallest rowSum or colSum, and let it be x. Place that number in the grid, and subtract x from rowSum and colSum. Continue until all the sums are satisfied. 相似题目 重构2 行二进制矩阵 中等...
建议和这一道题leetcode 480. Sliding Window Median 滑动窗口中位数 一起学习 代码如下: import java.util.Collections; import java.util.PriorityQueue; /* * 用一个最大堆存放比中位数小(或等于)的元素,用一个最小堆存放比中位数大(或等于)的元素。
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. Find the k pairs (u1,v1),(u2,v2) ...(uk,vk) with the smallest sums. ...
classSolution{private:structmycompare{booloperator()(pair<int,int>&p1,pair<int,int>&p2){returnp1.first+p1.second<p2.first+p2.second;}};public:vector<pair<int,int>>kSmallestPairs(vector<int>&nums1,vector<int>&nums2,intk){vector<pair<int,int>>res;priority_queue<pair<int,int>,vector...
Find the k pairs (u1,v1),(u2,v2) ...(uk,vk) with the smallest sums. 两个单调递增的整数数组,现分别从数组1和数组2中取一个数字构成数对,求找到k个和最小的数对。 思路 这题采用最大堆作为辅助的数据结构能够完美的解决我们的问题。观察数组我们可以看到,从nums1中任意取一个数字,其和nums2中的...
package leetcode import ( "container/heap" "sort" ) // 解法一 优先队列 func kSmallestPairs(nums1 []int, nums2 []int, k int) [][]int { result, h := [][]int{}, &minHeap{} if len(nums1) == 0 || len(nums2) == 0 || k == 0 { ...
简单利用数据结构 map<int, set<pair<int,int>>> ,遍历所有的数对可能性,取出前k个作为结果即可(map会自动按照第一个主参数排序)。 复杂度分析 时间复杂度:O(M * N) 空间复杂度:O(M * N) 代码 classSolution {public: vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2...