LeetCode Day36 Permutations II

在递归函数中要判断前面一个数和当前的数是否相等，如果相等，前面的数必须已经使用了，即对应的visited中的值为1，当前的数字才能使用，否则需要跳过

class Solution {
public:
    vector<vector<int> > permuteUnique(vector<int> &num) {
        vector<vector<int> > res;
        vector<int> out;
        vector<int> visited(num.size(), 0);
        sort(num.begin(), num.end());
        permuteUniqueDFS(num, 0, visited, out, res);
        return res;
    }
    void permuteUniqueDFS(vector<int> &num, int level, vector<int> &visited, vector<int> &out, vector<vector<int> > &res) {
        if (level >= num.size()) res.push_back(out);
        else {
            for (int i = 0; i < num.size(); ++i) {
                if (visited[i] == 0) {
                    if (i > 0 && num[i] == num[i - 1] && visited[i - 1] == 0) continue;
                    visited[i] = 1;
                    out.push_back(num[i]);
                    permuteUniqueDFS(num, level + 1, visited, out, res);
                    out.pop_back();
                    visited[i] = 0;
                }
            }
        }
    }
};

class Solution {
public:
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>> res;
        sort(nums.begin(),nums.end());
        DFS(nums,res,0);
        return res;
    }
    void DFS(vector<int>& nums,vector<vector<int>> & res,int start){
        if(start==nums.size()){res.push_back(nums);return;}
        else{
            for(int i=start;i<nums.size();i++){
                int j = i - 1;
                while (j >= start && nums[j] != nums[i]) j--;
                if (j == start - 1) {// if nums[i] does not equal to any vals in nums[idx, ... i - 1]
                swap(nums[start],nums[i]);
                DFS(nums,res,start+1);
                swap(nums[start],nums[i]);
                }            
            }
        }
    }
};