next_permutation

产生一个数列的全排列,自己想了一个实现算法,和stl里面的实现算法一样,具体的思路都在注释中。
代码中还实现了一种递归的做法,参考了网上的资料,递归的做法代码简短很多,但可操作性而言,没有next方式好。因为实际应用中需要针对每个permutation作相应的操作。用递归的话需要把操作写在permutation的实现中,耦合性很高。
另外,递归的方式受限于系统栈空间的大小,N只能在一定范围内处理。理论上,next的方式N可以无限大。

代码如下:

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#include <iostream>
#include <algorithm>
using namespace std;

#define N 5

int li[N];

bool next_permu(int start, int end){
if (end < start) return false;
// find the longest increase sub str from the end
int ptr1;
for (ptr1 = end - 1; ptr1 >= start; --ptr1){
if (li[ptr1] < li[ptr1 + 1])break;
}
if (ptr1 < start) return false;

// find the first one which is larger than li[ptr] from the end
// it is guaranteed that this ptr2 will be found
int ptr2;
for (ptr2 = end; ptr2 > ptr1; --ptr2){
if (li[ptr2] > li[ptr1])break;
}

// swap the numbers pointed by ptr1 and ptr2
int temp = li[ptr1];
li[ptr1] = li[ptr2];
li[ptr2] = temp;

// reverse the string from ptr1 + 1 to end
for (int i = ptr1 + 1, j = end; i < j; ++i, --j){
temp = li[i];
li[i] = li[j];
li[j] = temp;
}
return true;
}

// the recursive solution
void permu(int num){
if (num == 1){
for (int i = 0; i < N; ++i){
cout << li[i] << " ";
}
cout << endl;
return;
}
for (int i = N - num; i < N; ++i){
swap(li[N - num], li[i]);
permu(num - 1);
swap(li[N - num], li[i]);
}
}

int main(int argc, const char * argv[]) {
for (int i = 1; i <= N; ++i){
li[i - 1] = i;
}
do{
for (int i = 0; i < N; ++i){
cout << li[i] << " ";
}
cout << endl;
}
while (next_permu(0, N - 1));

permu(N);
return 0;
}

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