HDU steps 1.3.4 shǎ崽 OrOrOrOrz （int数组快排实现）

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shǎ崽 OrOrOrOrz

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3942 Accepted Submission(s): 1224

Problem Description

Acmer in HDU-ACM team are ambitious, especially shǎ崽, he can spend time in Internet bar doing problems overnight. So many girls want to meet and Orz him. But Orz him is not that easy.You must solve this problem first.
The problem is :
Give you a sequence of distinct integers, choose numbers as following : first choose the biggest, then smallest, then second biggest, second smallest etc. Until all the numbers was chosen .
For example, give you 1 2 3 4 5, you should output 5 1 4 2 3

Input

There are multiple test cases, each case begins with one integer N(1 <= N <= 10000), following N distinct integers.

Output

Output a sequence of distinct integers described above.

Sample Input

51 2 3 4 5

Sample Output

5 1 4 2 3

Author

WhereIsHeroFrom

Source

HDU女生专场公开赛——谁说女子不如男

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分析：这个题目基本思路先排序（注意由于测试数据较多，一定要使用快排，可以自己写，可以调用库函数），然后输出一个最大的、输出一个最小的一次输出。

代码：

import java.util.Arrays;import java.util.Scanner;public class Main {public static void main(String[] args) {int n;Scanner in = new Scanner(System.in);while (in.hasNext()) {n = in.nextInt();int[] a = new int[n];for (int i = 0; i < n; i++) {a[i]=in.nextInt();}Arrays.sort(a);//int f = 1;                //这种输出方法超时。 这个f（n）=n; 下面的输出方法 f(n)=n/2//int k=0,t=n-1;//for (int i = 0; i <n; i++) {//if(f==1){System.out.print(a[t--]);f=0;}//else {//System.out.print(a[k++]); k++; f=1;//}//if(i!=n-1){System.out.print(" ");}//}//System.out.println();if(n%2==0){int i=0,j=n-1;while(j-i>1){System.out.print(a[j--]+" "+a[i++]+" ");}System.out.print(a[j]+" "+a[i]);}else{int i=0,j=n-1;while(j>i){System.out.print(a[j--]+" "+a[i++]+" ");}System.out.print(a[j]);}System.out.println();}}}

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