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数组倒序排序

#include #include int isResult(int n,int number){ int sum = 0; while(number>0) { sum += pow(number%10,n); number/=10; } return sum;}int main(){ int n; scanf("%d",&n); int i = 0; for(i=pow(10,n-1);i

以下例为例 NSMutableArray *array = [[NSMutableArray alloc] initWithObjects:@"5",@"1",@"4",@"2",nil]; [array sortUsingSelector:@selector(compare:)]; NSEnumerator *enumerator = [array reverseObjectEnumerator]; [array release]; ar...

C语言程序: #include //冒泡排序void sort(double arr[], int n){int i, j;double temp;for(i=0; i

代码包含了正序和逆序: using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace ConsoleApplication4 { class Program { static void Main(string[] args) { string[] str = { "d","h","a","c","g"...

这个简单 String s[] = {1,2,3,4,5,6,7}; for (int i = 0; i < s.length; i ++){ System.out.print(s[s.length-1-i] +","); }

A = randint(1,10,[1 99])B = sort(A,'ascend')% 将 5:8 逆序排C = B;C(5:8) = sort(C(5:8),'descend')A = 2 74 45 93 47 42 84 52 21 67 B = 2 21 42 45 47 52 67 74 84 93 C = 2 21 42 45 74 67 52 47 84 93

void func(int * src, int *dst, int len) { int i; for(i = 0; i < len; i ++) dst[i] = src[len - i - 1]; }

import java.util.*; public class Ni { public static void main(String args[]){ Scanner sc=new Scanner(System.in); String str=sc.next(); for(int i=str.length()-1;i>=0;i--){ System.out.print(str.charAt(i)); } } }

首先阐述一下逆序对的概念。假设有一个数组为Array[0..n] 其中有元素a[i],a[j].如果 当ia[j],那么我们就称(a[i],a[j])为一个逆序对。 那么统计一个数组中的逆序对,有什么作用呢。逆序对可以反映插入排序的效率问题,如果逆序对数量多,那么...

public static List getArrayOrder(ArrayList array) { List imagesArray = new ArrayList(); String[] oArray = new String[array.size()]; //将排列好的数组循环加入到数组LIST中 for (int i = 0; i < oArray.length; i++) { imagesArray.add(...

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