【leetcode】Permutations

问题：

Given a collection of numbers, return all possible permutations.

For example,

[1,2,3]

have the following permutations:

[1,2,3]

,

[1,3,2]

,

[2,1,3]

,

[2,3,1]

,

[3,1,2]

,
and

[3,2,1]

.

分析：

permutations II 里的一样，这里排个序就很简单，一样的代码提交，同能通过，看来，考察的应该不是如此简单。
这里给个不排序的吧，其实还是用的排序的原理，只是嫁接了一层。

实现：

vector<vector<int> > permute(vector<int> &num) {
const int len = num.size();

vector<vector<int> > re;
re.push_back(num);
vector<int> index;

if(len <= 1)
return re;

int i = len - 1;
for (int n = 0; n < len; ++n)
{
index.push_back(n);
}
//reserve num, we don't plan to change it.
vector<int> cp(num);
while (1)
{
int ii = i;
--i;
if(i < 0 )
{
return re;
}
if(index[ii] > index[i]){
int j = len - 1;
while (index[j] < index[i]) --j;
swap(index[i], index[j]);
//you can write a reverse function yourself.
//Reverse(num, ii, len - 1);
reverse(index.begin() + ii,index.end());
for (int n = 0; n < len; ++n)
{
cp
= num[index
];
}
re.push_back(cp);
i = len - 1;
}

}
}