# 16 ms c++ solution, very simple idea, no recursion.

• At first, sort numbers into ascending order. Then use the idea of "Next Permutation", find the next permutation and next, next.... For example "2,1,3", -> 123 -> 132->..., until 321 that there is no number smaller than it's right side number.

Here is the code:

``````vector<vector<int>> permute(vector<int>& s)
{
vector<vector<int>> res;;
const int sz = s.size();
if (sz <= 1)
{
res.push_back(s);
return res;
}
sort(s.begin(), s.end());//先对数列进行排序
res.push_back(s);
int le;
int ri;
int cur;
int t;
while (1)
{
le = 0;
ri = sz - 1;
for (; ri > 0; --ri)
{
if (s[ri - 1] < s[ri])
break;
}
if (ri <= 0)//Already become a descending order, return.
{
return res;
}
cur = ri - 1;
le = ri;
ri = sz - 1;
for (; ri >= le && s[ri] <= s[cur]; --ri);
if (ri - cur)
{
t = s[ri];
s[ri] = s[cur];
s[cur] = t;
}
sort(s.begin() + cur+1, s.end());
res.push_back(s);
}
return res;
}``````

• PS: This solution can be used to solve "Permutations II".

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