c++ 實現部分排序，待改進

C++代碼（4）排列與組合

#include <assert.h>#include <iostream>#include <algorithm>using namespace std;template<typename T>void Perm(T* pT, int n, int m){    if (n == m)    {        for (int i=0; i<m; i++)            cout << pT[i] << " ";        cout << endl;        return;    }    else    {        for (int i=m; i<n; i++)        {            swap(pT[m], pT[i]);            Perm(pT, n, m+1);            swap(pT[m], pT[i]);        }    }}class Permutation{public:    enum {MAX_SIZE = 10};    Permutation(int nTotal, int nPart)    {        assert(0 < nTotal && nTotal < MAX_SIZE);        assert(0 < nPart && nPart <= nTotal);        m_nPart = (char)nPart;        m_nTotal = (char)nTotal;        for (char i=0; i<m_nTotal; i++)            m_Indexs[i] = i;    }    public:    bool ToNext();    char operator[] (int n)const    {        assert(0 <= n && n <= m_nTotal); // >??? 和原文不一致，需確認        return m_Indexs[n];    }private:    char m_Indexs[MAX_SIZE];    char m_nPart;    char m_nTotal;};bool Permutation::ToNext(){    if (m_nPart == m_nTotal)    {        // 為全排列        char nReverse = m_nTotal-1;          while (nReverse > 0 && m_Indexs[nReverse]<m_Indexs[nReverse-1])                 nReverse--;    // 尋找第一個要交換的元素           if (nReverse == 0)    // 找不到，表示全排已經完成              return false;                char nSwap = m_nTotal - 1;          while (m_Indexs[nSwap] < m_Indexs[nReverse-1])              nSwap--;    // 尋找第二個元素           swap(m_Indexs[nReverse-1], m_Indexs[nSwap]);    // 開始交換           reverse(m_Indexs+nReverse, m_Indexs+m_nTotal);    // 逆順           return true;    }    char nToSwap = m_nPart-1;    char nLeft = m_nTotal - 1;    if (m_Indexs[nLeft] > m_Indexs[nToSwap])    // 只改變尾數    {        while (nLeft > nToSwap && m_Indexs[nLeft]>m_Indexs[nToSwap])            nLeft--;            nLeft++;        swap(m_Indexs[nToSwap], m_Indexs[nLeft]);            return true;    }    while (nToSwap > 0 && m_Indexs[nToSwap]<m_Indexs[nToSwap-1] && m_Indexs[nToSwap]>m_Indexs[nLeft])        nToSwap--;    if (nToSwap == 0)    // 部分排列業已完成        return false;    nToSwap--;    // 已確定這個位置要參與交換了    if (m_Indexs[nToSwap] > m_Indexs[nLeft]) // 同參與部分排列的元素交換    {        char nReplace = m_nPart - 1;        while (m_Indexs[nReplace] < m_Indexs[nToSwap])            nReplace--;        swap(m_Indexs[nToSwap], m_Indexs[nReplace]);    }    else    // 同未參與部分排列的元素交換    {        while (nLeft >= m_nPart && m_Indexs[nLeft]>m_Indexs[nToSwap])            nLeft--;            nLeft++;        swap(m_Indexs[nToSwap], m_Indexs[nLeft]);            }    sort(m_Indexs+nToSwap+1, m_Indexs+m_nTotal);// 後面為剩下來的最小排列數    return true;}int main(){    const int N = 3, M = 3;    Permutation perm(N, M);    const char* sTests[N] = {"aaa", "bbb", "ccc"};    do    {        for (int i=0; i<N; i++)            cout << sTests[perm[i]] << " ";        cout << endl;    }while(perm.ToNext());    return 0;}