剑指offer38 字符串的排列

思路：

递归+回溯，关键是去重。

实现1（在原字符串中通过交换的方式生成排列）：

class Solution
{
public:
    void dfs(string &s, int p, vector<string>& res)
    {
        int n = s.length();
        if (p == n - 1) { res.push_back(s); return; }
        vector<bool> vis(26, false);
        for (int i = p; i < n; i++)
        {
            int tmp = s[i] - 'a';
            if (vis[tmp]) continue;
            vis[tmp] = true;
            swap(s[p], s[i]);
            dfs(s, p + 1, res);
            swap(s[p], s[i]);
        }
    }
    vector<string> permutation(string s)
    {
        vector<string> res;
        dfs(s, 0, res);
        return res;
    }
};

实现2（开辟新的缓冲区，不断选取未使用过的字符追加到缓冲区末尾生成排列）：

class Solution
{
public:
    void dfs(string &s, string& t, vector<bool>& vis, vector<string>& res)
    {
        int n = s.length();
        if (t.length() == n) { res.push_back(t); return; }
        for (int i = 0; i < n; i++)
        {
            if (vis[i]) continue;
            if (i > 0 and s[i - 1] == s[i] and !vis[i - 1]) continue;
            vis[i] = true;
            t.push_back(s[i]);
            dfs(s, t, vis, res);
            t.pop_back();
            vis[i] = false;
        }
    }
    vector<string> permutation(string s)
    {
        sort(s.begin(), s.end());
        int n = s.length();
        vector<bool> vis(n, false);
        vector<string> res;
        string t = "";
        dfs(s, t, vis, res);
        return res;
    }
};