时间复杂度是O(N*log(N)),空间复杂度是O(N). classSolution{public:vector<int>sortArray(vector<int>& nums){returnmergeSort(nums,0, nums.size()); }// sort nums[start, end)vector<int>mergeSort(vector<int>& nums,intstart,intend){if(start +1== end)returnvector<int>(1, nums[start]);...
Can you solve this real interview question? Sort an Array - Given an array of integers nums, sort the array in ascending order and return it. You must solve the problem without using any built-in functions in O(nlog(n)) time complexity and with the smal
Solution: first attempt: bubble sort. (Time out Exception) classSolution {publicint[] sortArray(int[] nums) {if(nums ==null||nums.length ==0){returnnull; }//bubble sort;for(inti = 0; i<nums.length-1;i++){for(intj = 0; j< nums.length-i-1; j++){if(nums[j]>nums[j+1])...
Can you solve this real interview question? GCD Sort of an Array - You are given an integer array nums, and you can perform the following operation any number of times on nums: * Swap the positions of two elements nums[i] and nums[j] if gcd(nums[i], nu
题目地址:https://leetcode.com/problems/sort-array-by-parity-ii 题目描述 Given an array A of non-negative integers, half of the integers in A are odd, and half of the integers are even. Sort the array so that whenever A[i] is odd, i is odd; and whenever A[i] is even, i is ...
leetcode 912 排序数组 sort-an-array【ct】,思路:快速排序soeasy一次通过啦
Given an array A of non-negative integers, return an array consisting of all the even elements of A, followed by all the odd elements of A. You may return any answer array that satisfies this condition. Example 1: Input: [3,1,2,4] ...
今天介绍的是LeetCode算法题中Easy级别的第258题(顺位题号是1122)。给定两个数组arr1和arr2,arr2中的元素是不同的,arr2中的所有元素也在arr1中。 对arr1的元素进行排序,使arr1中元素的相对顺序与arr2中的相对顺序相同。未出现在arr2中的元素应按升序放置在arr1的末尾。
Leetcode Sort Array ①归并排序 归并排序(MERGE-SORT)是建立在归并操作上的一种有效的排序算法,该算法是采用分治法(Divide and Conquer)的一个非常典型的应用。将已有序的子序列合并,得到完全有序的序列;即先使每个子序列有序,再使子序列段间有序。若将两个有序表合并成一个有序表,称为二路归并。
classSolution:defmerge_sort(self,nums):# If the array length is less than or equal to 1, return the array (base case for recursion)iflen(nums)<=1:returnnums# Find the middle pointmid=len(nums)//2# Recursively sort the left halfleft_half=self.merge_sort(nums[:mid])# Recursively sort...