算法程序题：(转自Csdn)

Question

算法程序题：    该公司笔试题就1个，要求在10分钟内作完。    题目如下：用1、2、2、3、4、5这六个数字,打印出所有不同的排列，如：512234、412345等，要求："4"不能在第三位，"3"与"5"不能相连。

Answer 1

using System;
using System.Collections.Generic;
using System.Text;namespace Test
{
    class Program
    {
        public static void Rotate(int[] arr, int bound) //旋转递归法，产生全排列
        {
            int len = arr.Length;
            if (bound > len)
            {
                throw new Exception("Invalid Argument.");
            }
            if (bound < 2)
            {
                if (arr[2] != 4) //第三位不为4
                {
                    int sum = 0;
                    int ori = 0;
                    foreach (int i in arr)
                    {
                        sum += i;
                        sum *= 10;
                        if(ori==3&&i==5 ||ori==5&&i==3) //35相连
                        {
                                sum = 0;
                                break;
                        }
                        ori = i;
                    }
                    if (sum > 0)
                    {
                        Console.WriteLine(sum/10);//去掉最后的0
                    }                }
            }
            else
            {
                for (int i = 0; i < bound; ++i)
                {
                    Rotate(arr, bound - 1); //递归旋转剩下部分
                    int k = len - bound;
                    int temp = arr[k];
                    for (; k < len; k++) //旋转数组
                    {
                        if (k == len - 1)
                        {
                            arr[k] = temp;
                        }
                        else
                        {
                            arr[k] = arr[k + 1];
                        }
                    }
                }
            }
        }
        public static void Main()
        {
            const int SIZE = 6;
            int[] arr = new int[SIZE] { 1, 2, 2, 3, 4, 5 };
            Array.Sort(arr);
            Console.WriteLine("Begin...");
            Rotate(arr, SIZE);
        }    }
}

Answer 2

#region 
//用循环实现
int[] num = { 1, 2, 2, 3, 4 };
int index1, index2, index3, index4;
for (int i = 0; i < 5; i++)
{
    for (int i1 = 1; i1 <= 4; i1++)
    {
        index1 = (i + i1)%5;
        for (int i2 = 1; i2 <= 3; i2++)
        {
            index2 =(index1 +i2)%5;
            //判断下标是否出现过
            while (index2 == i || index2 == index1)
            {
                index2++;
                index2 %= 5;
            }
            index3 = index2;
            for (int i3 = 1; i3 <= 2; i3++)
            {
                //判断下标是否出现过
                while (index3 == i ||index3 == index1 || index3 == index2)
                {
                    index3++;
                    index3 %= 5;
                }
                index4 = (index3 + 1)%5;
                //判断下标是否出现过
                while (index4 == i || index4 == index1 
                    || index4 == index2 || index4 == index3)
                {
                    index4++;
                    index4 %= 5;
                }                Console.Write("{0} {1} {2} {3} {4}", i, index1,index2,index3,index4);
                
                Console.WriteLine("\t{0} {1} {2} {3} {4}"
                    ,num[i], num[index1],num[index2],num[index3],num[index4]);
            }
        }
    }
}
#endregion

Answer 3

to fd7893(看着办吧) ( ) 
如果数组大小是100个呢？
写100个嵌套循环？呵呵

Answer 4

学习了..
不过shrinerain(圣影雨)的算法未排除重复数据.......

Answer 5

To: shrinerain(圣影雨) ( ) 
如果数组大小不确定，当然要选择递归。这毫无疑问！

Answer 6

#region
            //用循环实现
            int[] num = { 1, 2, 2, 3, 4 };
            int index1, index2, index3, index4;
            for (int i = 0; i < 5; i++)
            {
                for (int i1 = 1; i1 <= 4; i1++)
                {
                    index1 = (i + i1) % 5;
                    for (int i2 = 1; i2 <= 3; i2++)
                    {
                        index2 = (index1 + i2) % 5;
                        //判断下标是否出现过
                        while (index2 == i || index2 == index1)
                        {
                            index2++;
                            index2 %= 5;
                        }
                        index3 = index2;
                        for (int i3 = 1; i3 <= 2; i3++)
                        {
                            //判断下标是否出现过
                            while (index3 == i || index3 == index1 || index3 == index2)
                            {
                                index3++;
                                index3 %= 5;
                            }
                            index4 = (index3 + 1) % 5;
                            //判断下标是否出现过
                            while (index4 == i || index4 == index1
                                || index4 == index2 || index4 == index3)
                            {
                                index4++;
                                index4 %= 5;
                            }                            Console.WriteLine("{0} {1} {2} {3} {4}\t{5} {6} {7} {8} {9}", i, index1, index2, index3, index4
                                        , num[i], num[index1], num[index2], num[index3], num[index4]);
                            index3 = (index3 + 1) % 5;
                        }
                        index2 = (index2 + 1) % 5;
                    }
                    index1 = (index1 + 1) % 5;
                }
            }
            #endregion
循环结束时忘记累加了。