[LeetCode] Permutations

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]

.

思路：回溯法，排序树，题目没有考虑到元素有重复的情况，我再最后加了sort+unique

有关回溯法 子集树和排序树 /article/6745870.html

class Solution {
vector<vector<int> > m_result;
public:
void swap(int & x, int & y)
{
int tmp = x;
x = y;
y = tmp;
}
void dfs(vector<int> & num, int dep )
{
if(dep == num.size())
{
m_result.push_back(num);
return;
}

for(int i = dep; i < num.size(); i++)
{
//if(i != dep && num[i] == num[dep])
//    continue;
swap(num[i], num[dep]);
dfs(num, dep + 1);
swap(num[i], num[dep]);
}
}

vector<vector<int> > permute(vector<int> &num)
{
dfs( num, 0);

//erase the duplicate
//sort(m_result.begin(), m_result.end());
//m_result.erase(unique(m_result.begin(), m_result.end()), m_result.end());

return m_result;
}
};

思路2： 从 http://www.cnblogs.com/remlostime/archive/2012/11/13/2767818.html 看来的想法，用canUse标记是否使用，没用才可以用，否则就不能用。

为什么用这个呢？因为这种思路对Permutations II 非常有帮助。

class Solution {
private:
vector<vector<int> > ret;
bool canUse[100];
int a[100];
public:
void dfs(int dep, int maxDep, vector<int> &num)
{
if (dep == maxDep)
{
vector<int> ans;
for(int i = 0; i < maxDep; i++)
ans.push_back(a[i]);
ret.push_back(ans);
return;
}

for(int i = 0; i < maxDep; i++)
if (canUse[i])
{
canUse[i] = false;
a[dep] = num[i];
dfs(dep + 1, maxDep, num);
canUse[i] = true;
}
}

vector<vector<int> > permute(vector<int> &num) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
ret.clear();
memset(canUse, true, sizeof(canUse));
dfs(0, num.size(), num);
return ret;
}
};