Permutations- LeetCode

Given a collection of distinct integers, return all possible permutations.

Example:

Input: [1,2,3]
Output:
[
  [1,2,3],
  [1,3,2],
  [2,1,3],
  [2,3,1],
  [3,1,2],
  [3,2,1]
]

STL：

class Solution {
public:
    vector<vector<int>>ans;
    vector<vector<int>> permute(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        do {
            ans.push_back(nums);
        } while (next_permutation(nums.begin(), nums.end()));
        return ans;
    }
};

DFS动态数组：

class Solution {
public:
    vector<vector<int>> ans;
    vector<int> path;
    vector<bool> vis;
    
    vector<vector<int>> permute(vector<int>& nums) {
        vis = vector<bool>(nums.size(), false);
        dfs(nums, 0);
        return ans;
    }
    
    void dfs(vector<int>& nums, int step) {
        if (step == nums.size()) {
            ans.push_back(path);
            return ;
        }
        for (int i = 0; i < nums.size(); i ++ ) {
            if(!vis[i]){
                vis[i] = true, path.push_back(nums[i]);
                dfs(nums, step + 1);
                vis[i] = false, path.pop_back();
            }
        }
    }
};

DFS普通数组：

class Solution {
public:
    vector<vector<int>> ans;
    vector<int> path;
    vector<bool> vis;
    
    vector<vector<int>> permute(vector<int>& nums) {
        vis = vector<bool>(nums.size(), false);
        path = vector<int>(nums.size());
        dfs(nums, 0);
        return ans;
    }
    
    void dfs(vector<int>& nums, int step) {
        if (step == nums.size()) {
            ans.push_back(path);
            return ;
        }
        for (int i = 0; i < nums.size(); i ++ ) {
            if(!vis[i]){
                vis[i] = true, path[i] = nums[step];
                dfs(nums, step + 1);
                vis[i] = false;
            }
        }
    }
};