Given an unsorted array nums, reorder it such that nums[0] < nums[1] > nums[2] < nums[3].... Example 1: Input: nums = [1, 5, 1, 1, 6, 4] Output: One possible answer is [1, 4, 1, 5, 1, 6]. Example 2: Input: nums = [1, 3, 2, 2, 3, 1] Output: One po
Given an unsorted array nums, reorder it in-place such that nums[0] <= nums[1] >= nums[2] <= nums[3].... For example, given nums = [3, 5, 2, 1, 6, 4], one possible answer is [1, 6, 2, 5, 3, 4]. 这道题让我们求摆动排序,跟Wiggle Sort II相比起来,这道题的条件宽松很多,只因为多了一个等号
//三向切分的快速排序 //这种切分方法对于数组中有大量重复元素的情况有比较大的性能提升 public static void main(String[] args) { Scanner input = new Scanner(System.in); int n = input.nextInt(); int[] a = new int[n]; for(int i = 0; i < a.length; i++) a[i] = (int)(Math.random() * 100); //System
Given an unsorted array nums, reorder it in-place such that nums[0] <= nums[1] >= nums[2] <= nums[3].... For example, given nums = [3, 5, 2, 1, 6, 4], one possible answer is [1, 6, 2, 5, 3, 4]. 给一个没有排序的数组,将其重新排序成nums[0] <= nums[1] >= nums[2
数据结构实验之排序三:bucket sort Time Limit: 250 ms Memory Limit: 65536 KiB Problem Description 根据人口普查结果,知道目前淄博市大约500万人口,你的任务是帮助人口普查办公室按年龄递增的顺序输出每个年龄有多少人,其中不满1周岁的按0岁计算,1到2周岁的按1岁计算,依次类推,大于等于100岁的老人全部按100岁计算. Input 输入第一行给出一个正整数N(<=5000000),随后连续给出N个整数表示每个人的年龄,数字间
Given an unsorted array nums, reorder it such that nums[0] < nums[1] > nums[2] < nums[3].... Example:(1) Given nums = [1, 5, 1, 1, 6, 4], one possible answer is [1, 4, 1, 5, 1, 6]. (2) Given nums = [1, 3, 2, 2, 3, 1], one possible answer is [2, 3
Given an unsorted array nums, reorder it such that nums[0] < nums[1] > nums[2] < nums[3].... Example 1: Input: nums = [1, 5, 1, 1, 6, 4] Output: One possible answer is [1, 4, 1, 5, 1, 6]. Example 2: Input: nums = [1, 3, 2, 2, 3, 1] Output: One po
快速排序 快速排序通过一个切分元素将数组分成两个子数组,左子数组小于等于切分元素,右子数组大于切分元素,将这两个子数组排序,也就是将整个数组排序了. 代码如下: public class Sort{ public static void quickSort(int []arr){ if(arr=null||arr.length<2) return; quickSort(arr,0,arr.length-1); } public static void quickSort(int[]arr,i
