算法(全排列算法封装)

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本算法是教材中的全排列方法之一,本人仅做封装,在此感谢发现算法和传播算法的大牛们.
    /// <summary>    /// 全排列算法,算法原理:Perm(n)=[n]*Pern(n-1).N的全排列等于将N个数取一个放在第N个位置后,剩下的N-1个数做全排列。    /// 这个算法的一个用途是进行行列式的展开和计算,这也是这次封装这个算法的目的。    /// </summary>    public class Permulation    {        /// <summary>        /// 排列结果        /// </summary>        private List<List<int>> _PermArray { get; set; }        /// <summary>        /// 去重用.        /// </summary>        private Dictionary<string, List<int>> _NoRepeatArray { get; set; }        /// <summary>        /// 要排列的整数数组        /// </summary>        private int[] _Numbers { get; set; }        /// <summary>        /// 元素个数        /// </summary>        private int _N { get; set; }        /// <summary>        /// 是否去重.        /// </summary>        private bool _RemoveDup { get; set; }                /// <summary>        /// 全排列计数        /// </summary>        public int TotalCount { get; set; }        /// <summary>        /// 排列结果        /// </summary>        public List<List<int>> PermulationArray        {            get            {                return _PermArray;            }        }        /// <summary>        /// 任意给定数字数组进行全排列        /// </summary>        /// <param name="Numbers">数组</param>        /// <param name="RemoveDup">是否去重</param>        public Permulation(int[] Numbers, bool RemoveDup = false)        {            _NoRepeatArray = new Dictionary<string, List<int>>();            _PermArray = new List<List<int>>();            TotalCount = 0;            _Numbers = Numbers;            _N = Numbers.Count();            _RemoveDup = RemoveDup;        }        /// <summary>        /// 自然数1-N全排列        /// </summary>        /// <param name="N"></param>        public Permulation(int N)        {            _PermArray = new List<List<int>>();            TotalCount = 0;            _Numbers = new int[N];            for (int i = 1; i <= N; i++)            {                _Numbers[i - 1] = i;            }            _N = N;        }        /// <summary>        /// 交换位置.        /// </summary>        /// <param name="Nums"></param>        /// <param name="i"></param>        /// <param name="j"></param>        private void Swap(int[] Nums, int i, int j)        {            int theTemp = Nums[i - 1];            Nums[i - 1] = Nums[j - 1];            Nums[j - 1] = theTemp;        }        /// <summary>        /// 执行全排列        /// </summary>        public void DoCalculation()        {            DoArray(1);        }        /// <summary>        /// 递归算法进行全排列.        /// </summary>        /// <param name="NextIndex"></param>        private void DoArray(int NextIndex)        {            if (NextIndex > _N)            {                var theNums = new List<int>();                //利用字典本身的字符串哈希算法判重。                var theSeqStr = "";                for (int i = 0; i < _N; i++)                {                    //注意这里需要分割,防止1 23和12 3之类造成的重复.                    theSeqStr += "," + _Numbers[i];                    theNums.Add(_Numbers[i]);                }                if (_RemoveDup)                {                    if (!_NoRepeatArray.ContainsKey(theSeqStr))                    {                        _NoRepeatArray.Add(theSeqStr, theNums);                        _PermArray.Add(theNums);                        TotalCount++;                    }                }                else                {                    _PermArray.Add(theNums);                    TotalCount++;                }            }            else            {                //与后面的所有位置进行交换,但注意,每次交换完,应复原。                for (int i = NextIndex; i <= _N; i++)                {                    Swap(_Numbers, NextIndex, i);                    DoArray(NextIndex + 1);                    //复原                    Swap(_Numbers, NextIndex, i);                }            }        }    }


注:本算法只经过简单测试,没经过大批量测试。

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