5个 数 3个 一组 求10组不重复的算法

Question

最简单的,不考虑效率
 string[] number ={ "1","2","3","4","5"};            int maxNum = 10;            List<string> ret = new List<string>();            for (int i = 0; i < 5; i++)
            {
                string temp = number[i];
                for (int j = 0; j < 5; j++)
                {
                    for (int n = 0; n < 5; n++)
                    {                        
                        if(j!=i && n!=i && n!=j)
                        {
                            ret.Add(number[i] + number[j] + number[n]);
                            if (ret.Count >= maxNum)
                                break;
                        }
                        if (ret.Count >= maxNum)
                            break;
                    }
                    if (ret.Count >= maxNum)
                        break;
                }
                if (ret.Count >= maxNum)
                    break;            }            string outStr = string.Empty;
            foreach (string s in ret)
            {
                if (!string.IsNullOrEmpty(outStr))
                    outStr += ",";
                outStr += s;            }            MessageBox.Show(outStr);

Answer 1

看了LS各位高手的说法，我感觉LZ的意思是那5个数里 本身就存在重复的。如果5个数都不重复的话，就是60种了。所以先选出不重复的数，再输出排列的组合。

Answer 2

//选出不重复的数
        private List<int> SelectNum(List<int> nums)
        {
            for (int i = 0; i < nums.Count; i++)
            {
                for (int j = i + 1; j < nums.Count; j++)
                {
                    if (nums[i] == nums[j])
                    {
                        nums.Remove(nums[i]);
                        i = -1;
                        break;
                    }
                }
            }
            return nums;
        }//测试
static void main(string[] args)
{
            List<int> nums = new List<int>();            //3、5、9、7
            nums.Add(3);
            nums.Add(5);
            nums.Add(9);
            nums.Add(7);
            nums.Add(3);            nums = SelectNum(nums);            string newStr = "";            for (int i = 0; i < nums.Count; i++)
            {
                newStr += nums[i].ToString();
                for (int j = 0; j < nums.Count; j++)
                {
                    if (i == j)
                    {
                        continue;
                    }
                    else
                    {
                        newStr += nums[j].ToString();
                        for (int z = 0; z < nums.Count; z++)
                        {
                            if (z == i || z == j)
                            {
                                continue;
                            }
                            else
                            {
                                newStr += nums[z].ToString();                                if (newStr.Length == 3)
                                    MessageBox.Show(newStr);
                                newStr = newStr.Substring(0, 2);
                            }
                        }
                        newStr = newStr.Substring(0, 1);
                    }
                }
                newStr = "";
            }
}