Leetcode - Permutations I，II

Leetcode - 046 Permutations

全排列问题是回溯的典型例题：
1.可行解的组成形式是给定数组中的所有数的组合，故而大小上可以作为可行解判定条件
2.每次需要在剩下可被选中的集合中选择一个，创建mask数组

class Solution {
public:

    void dfs(vector> &vct, vector &cur, vector& nums,vector & used)
    {
        if (cur.size() == nums.size())
        {
            vct.push_back(cur);
            return;
        }
        for (int i = 0; i < nums.size(); ++i)
        {
            if (used[i] == 0)
            {
                cur.push_back(nums[i]);
                used[i] = 1;
                dfs(vct, cur, nums, used);
                used[i] = 0;
                cur.pop_back();
            }
        }
    }

    vector> permute(vector& nums) {
        vector> vct;
        int n = nums.size();
        if (n <= 0)
            return vct;
        vector cur;
        vector used(n, 0);
        dfs(vct, cur, nums, used);
        return vct;
    }
};

Leetcode - 047. Permutations II

diff ： 需要考虑val1 = val2 的情况，需要sort将相同元素聚类，然后可以参考前文 Leetcode - 040. Combination Sum II 去重的方法

class Solution {
public:

    void dfs(vector> &vct, vector &cur, vector& nums, vector & used)
    {
        if (cur.size() == nums.size())
        {
            vct.push_back(cur);
            return;
        }
        for (int i = 0; i < nums.size(); ++i)
        {
            if (used[i] == 0)
            {
                int pre_index = i - 1;
                bool repeated = false;
                while (pre_index >= 0 && nums[pre_index] == nums[i])
                {
                    if (used[pre_index] == 0)
                    {
                        repeated = true;
                        break;
                    }
                    --pre_index;
                }
                if (repeated)
                    continue;
                cur.push_back(nums[i]);
                used[i] = 1;
                dfs(vct, cur, nums, used);
                used[i] = 0;
                cur.pop_back();
            }
        }
    }

    vector> permuteUnique(vector& nums) {
        vector> vct;
        vector cur;
        int n = nums.size();
        if (n <= 0)
            return vct;
        vector used(n, 0);
        dfs(vct, cur, nums, used);
        return vct;
    }
};