参考: 花花酱 LeetCode 912. Sort an Array解法一:快速排序 时间复杂度: O(nlogn) ~ O(n^2) 空间复杂度:O(logn) ~ O(n) class Solution { public: vector<int> sortArray(vector<int>& nums) { sort(nums , 0 , nums.size() - 1); return nums; } //快速排序 //左闭右闭 void sort...
class Solution: def merge_sort(self, nums): # If the array length is less than or equal to 1, return the array (base case for recursion) if len(nums) <= 1: return nums # Find the middle point mid = len(nums) // 2 # Recursively sort the left half left_half = self.merge_sort...
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]);...
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])...
class Solution { public: vector<int> sortArray(vector<int>& nums) { sort(nums.begin(), nums.end()); return nums; } }; 1. 2. 3. 4. 5. 6. 7. 桶排序 桶排序就是遍历所有元素,把元素的个数累加到对应的桶上,最后进行一次遍历把统计的数字放到结果中即可。
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
[LeetCode] 912. Sort an Array Given an array of integersnums, sort the array in ascending order. Example 1: AI检测代码解析 Input: nums = [5,2,3,1] Output: [1,2,3,5] 1. 2. Example 2: AI检测代码解析 Input: nums = [5,1,1,2,0,0]...
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
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] ...
215. Kth Largest Element in an Array quicksort就是选出某一个数在数组里面应该待的位置。上一篇已经提到了。所以这里就可以用quick sort来找第n-k的位置上的数。n为数组长度。 my solution: class Solution { public int findKthLargest(int[] nums, int k) { ...