题目：给定1-n数字，排列组合。
解法：递归。第一个数字有n种选择，第二个数字有n-1种选择，依次递归排列输出。用数组表示n个数字，用过的数字置0。
实现语言：C++
#include <iostream>
using namespace std; /************************************************************************/
/*
num : 需要排列的数组
count : 数组总数
numC： 已经排列的数组
iUse：已经排列的个数
iNull：置0的个数
sum: 总排列数
*/
/************************************************************************/
template <class T>
void ComBineNum(T *num, const int count, T *numC, int iUse, int* sum)
{
int iNull = 0;
T *newNum = new T[count];
for (int i = 0; i < count; ++i){
memcpy(newNum, num, count);
if (newNum[i] == 0){
++iNull;
if (iNull == count){
for (int i = 0; i < count; ++i){
cout << numC[i];
}
cout << endl;
++(*sum);
return;
}
continue;
}
numC[count - iUse] = newNum[i];
newNum[i] = 0;
ComBineNum(newNum, count, numC, iUse - 1, sum);
}
delete[] newNum;
} int main()
{
int sum = 0;
const int count = 4;
char num[count], pNum[count];
for (int i = 0; i < count; ++i){
num[i] = i + '1';
}
ComBineNum<char>(num, count, pNum, count, &sum);
cout << "sum :" << sum << endl; sum = 1;
for (int i = 1; i <= count; ++i){
sum *= i;
}
cout << "sum :" << sum << endl; return 0;
}
输出：
template <class T>
void Swap(T& a, T& b)
{
T c = a;
a = b;
b = c;
} template <class T>
void Perm(T list[], int k, int m, int* count)
{
if (k == m){
for (int i = 0; i < m; ++ i){
cout << list[i];
}
cout << endl;
++(*count);
}
else{
for (int i = k; i < m; ++i){
Swap(list[k], list[i]);
Perm(list, k + 1, m, count);
Swap(list[i], list[k]);
}
}
} int main()
{
const int m = 4;
int count = 0;
int list[m];
for (int i = 0; i < m; ++i){
list[i] = i + 1;
} Perm(list, 0, m, &count);
cout << count;
return 0;
}
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