Given an array of 2n integers, your task is to group these integers into n pairs of integer, say (a1, b1), (a2, b2), ..., (an, bn) which makes sum of min(ai, bi) for all i from 1 to n as large as possible. Example 1: Input: [1,4,3,2] Output: 4 Explanation: n i...
题目给出一个含2n个整数的数组,要求拆成n对整数对,使得每个整数对的最小值的和最大,并返回这个最大值。可以将数组从小到大进行排序,然后两两组合,这样可以得到最大值。 python代码 class Solution: def arrayPairSum(self, nums): """ :type nums: List[int] :rtype: int """ nums.sort() newnum...
leetcode array专题笔记 以前也刷leetcode,但都不够系统,基本上是今天一榔头明天一棒槌,今天计划开始系统性地刷题,在查漏补缺的同时,也培养自己开始记笔记的习惯。以array开始,立下flag,每周更新一次,把比较有意思的题目记录下来。(2020.06.10 updated,终于结束了996) easy: 1266给定遍历顺序和运动规则,计算最小遍...
leetcode 561. Array Partition I 数组元素分组的最小值的和的最大值+直接排序即可,Givenanarrayof2nintegers,yourtaskisllifrom1tonaslargeaspossib...
参考解答1:使用快速排序的 partition 的思想完成。 public class Solution2 { private static Random random = new Random(System.currentTimeMillis()); public int findKthLargest(int[] nums, int k) { int len = nums.length; if (len == 0 || k > len) { ...
Note that the explanation of this method under the solution section on Leetcode is wrong: It has nothing to do with whether n%k = 0 or not. Though the issue will occur when n%k = 0, it is because n and k are coprime, not n%k = 0. For instance, n = 6, k = 4, n%k =...
https://leetcode-cn.com/problems/array-partition-i/description/ 要求 给定长度为 2n 的数组, 你的任务是将这些数分成 n 对, 例如 (a1, b1), (a2, b2), ..., (an, bn) ,使得从1 到 n 的 min(ai, bi) 总和最大。 输入:[ 1,4,3,2]输出:4解释:n 等于 ...
https://leetcode-cn.com/submissions/detail/6613451/ 数组的分组,这道题也是开始有点唬人,但其实想明白了很简单 将数组两两分组,取出每个分组里较小的那个,然后求和,需要让这个和最大。 那么举个例子 [1,444,888,3] 这里,如果把1和444分成一组,888和3分成一组,那么这个和只能是4,但如果是把444和888放到...
arr, partition the array into (contiguous) subarrays of lengthat mostk. After partitioning, each subarray has their values changed to become the maximum value of that subarray. Returnthe largest sum of the given array after partitioning. Test cases are generated so that the answer fits in a32...
class Solution { public: vector<int> productExceptSelf(vector<int>& nums) { if (nums.size()<=1) { return nums; } vector<int> res(nums.size(),1); long long product = 1; int zeroNum = 0; for (int i = 0; i < nums.size(); i++) ...