https://github.com/grandyang/leetcode/issues/912 类似题目: Kth Largest Element in an Array Sort Colors Count of Range Sum 参考资料: https://leetcode.com/problems/sort-an-array/ https://leetcode.com/problems/sort-an-ar
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...
vector<int> B =mergeSort(nums, start + L /2, end);returnmerge(A, B); }// merge two sorted arrayvector<int>merge(vector<int>& A, vector<int>& B){intM = A.size(), N = B.size();if(M ==0)returnB;if(N ==0)returnA; vector<int> res;autoita = A.begin();autoitb = ...
Given an array of integers nums, sort the array in ascending order. Example 1: Input: [5,2,3,1] Output: [1,2,3,5] 1. 2. Example 2: Input: [5,1,1,2,0,0] Output: [0,0,1,1,2,5] 1. 2. Note: 1 <= A.length <= 10000 -50000 <= A[i] <= ...
Sort an Array quicksort1: quicksort2: quicksort3:...leetcode 912. 排序数组 这道题就是纯粹的数组排序问题,救助这道题复习一下各种排序算法。 选择排序 这种排序方式应该是很简单的,容易理解,对于长度为n的数组,遍历n-1次,每次在无序的数组段中选择最小的一个元素和无序数组第一个元素交换位置。
an empty array for the sorted elements sorted_array = [] # Initialize pointers for both halves i = j = 0 # Traverse both arrays and in each iteration add the smaller element to the sorted array while i < len(left) and j < len(right): if left[i] < right[j]: sorted_array.append...
[LeetCode] 912. Sort an Array Given an array of integersnums, sort the array in ascending order. Example 1: Input: nums = [5,2,3,1] Output: [1,2,3,5] 1. 2. Example 2: Input: nums = [5,1,1,2,0,0] Output: [0,0,1,1,2,5]...
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
922. Sort Array By Parity II # 题目 # 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 even. You may
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