[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 http://www.cnblogs.com/diegodu/p/3805669.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;
    }
};