【leetcode】Permutations (middle)

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].

求没有重复数字的全排列

思路：用的标准回溯法，一次AC

class Solution {
public:
    vector<vector<int> > permute(vector<int> &num) {
        vector<vector<int>> ans;
        if(num.empty())
        {
            return ans;
        }

        vector<int> X(num.size());
        vector<vector<int>> S(num.size());
        int k = ;
        S[k] = num;

        while(k >= )
        {
            while(!S[k].empty())
            {
                X[k] = S[k].back();
                S[k].pop_back();
                if(k < num.size() - )
                {
                    k++;
                    S[k] = num;
                    for(int i = ; i < k; i++)
                    {
                        vector<int>::iterator  it;
                        if((it = find(S[k].begin(), S[k].end(), X[i])) != S[k].end())
                        {
                            S[k].erase(it);
                        }
                    }
                }
                else
                {
                    ans.push_back(X);
                }
            }
            k--;
        }

        return ans;
    }
};

网上也有用递归的

class Solution {
public:
    vector<vector<int> > permute(vector<int> &num) {
        vector<vector<int> > result;

        permuteRecursive(num, , result);
        return result;
    }

    // permute num[begin..end]
    // invariant: num[0..begin-1] have been fixed/permuted
    void permuteRecursive(vector<int> &num, int begin, vector<vector<int> > &result)    {
        if (begin >= num.size()) {
            // one permutation instance
            result.push_back(num);
            return;
        }

        for (int i = begin; i < num.size(); i++) {
            swap(num[begin], num[i]);
            permuteRecursive(num, begin + , result);
            // reset
            swap(num[begin], num[i]);
        }
    }
};