目标:实现1,2,3的全排列
方法1:dfs
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;
const int N = 1e3 + 10;
int n, a[N], book[N] ,b[N];
//dfs模板
void dfs(int n, int id)
{
if (id > n)
{
for (int i = 1; i <= n; i++)
{
cout << b[i] << " ";
}
cout << endl;
}
for (int i = 1; i <= n; i++)
{
if (!book[i])
{
b[id] = a[i];
book[i] = 1;
dfs(n, id + 1); //深度搜索
book[i] = 0; //恢复标记
}
}
}
//全排列
int main()
{
cin >> n;
for (int i = 1; i <= n; i++)
cin >> a[i];
dfs(n, 1);
return 0;
}
方法2: 使用next_permutation函数
作用:生成下一个排列,例如:1,2,3的下一个是1,3,2。与之对应的prev_permutation是生成上一个全排列。
要点:
- 包含头文件<algorithm>
- 将原来的序列先升序,否则无法获得所有全排列
- 不会产生重复序列
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
const int N = 1e3 + 10;
int n, a[N];
int main()
{
//普通数组
cin >> n;
for (int i = 1; i <= n; i++)
cin >> a[i];
//dfs(n, 1);
do
{
for (int i = 1; i <= n; i++)
{
cout << a[i]<<" ";
}
cout << endl;
} while (next_permutation(a+1, a + n + 1));
//vector容器
vector<int> c = { 2,1,3 };
sort(c.begin(), c.end());
do
{
for (int i = 0; i < c.size() ; i++) //注意如何遍历的
{
cout << c[i] << " ";
}
cout << endl;
} while (next_permutation(c.begin(), c.end())); //注意!!!!!!
return 0;
}
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