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