【Sword Finger 38】Character arrangement

Method 1: dfs + pruning: time O( $n!$ ), space O(n)

Problem-solving ideas:

• Full arrangement of strings, insert the structure of each different arrangement into the array
• Loop through the array through dfs + and swap element positions to achieve full arrangement
• During the full arrangement process, multiple elements in the array may occupy the same position repeatedly, resulting in duplicate results
• Therefore, a fixed-size array is used, and the position of each row corresponds to the subscript ++

class Solution {
    
    
public:
    vector<string> ans;
    vector<string> permutation(string s) 
    {
    
    
        dfs(s, 0);
        return ans;
    }
    void dfs(string& s, int pos)
    {
    
    
        // 直接对 s 进行全排列
        if (pos == s.size() - 1)
        {
    
    
            ans.push_back(s);
            return;
        }
        int board[1024] = {
    
     0 };
        for (int i = pos; i < s.size(); ++i)
        {
    
    
            if (board[s[i]] != 0)
                continue;
            board[s[i]]++;
            swap(s[i], s[pos]);
            dfs(s, pos + 1);
            swap(s[i], s[pos]);
        }
    }
};