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）：
1 class Solution 2 { 3 public: 4 void dfs(string &s, int p, vector<string>& res) 5 { 6 int n = s.length(); 7 if (p == n - 1) { res.push_back(s); return; } 8 vector<bool> vis(26, false); 9 for (int i = p; i < n; i++) 10 { 11 int tmp = s[i] - 'a'; 12 if (vis[tmp]) continue; 13 vis[tmp] = true; 14 swap(s[p], s[i]); 15 dfs(s, p + 1, res); 16 swap(s[p], s[i]); 17 } 18 } 19 vector<string> permutation(string s) 20 { 21 vector<string> res; 22 dfs(s, 0, res); 23 return res; 24 } 25 };
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）：
1 class Solution 2 { 3 public: 4 void dfs(string &s, string& t, vector<bool>& vis, vector<string>& res) 5 { 6 int n = s.length(); 7 if (t.length() == n) { res.push_back(t); return; } 8 for (int i = 0; i < n; i++) 9 { 10 if (vis[i]) continue; 11 if (i > 0 and s[i - 1] == s[i] and !vis[i - 1]) continue; 12 vis[i] = true; 13 t.push_back(s[i]); 14 dfs(s, t, vis, res); 15 t.pop_back(); 16 vis[i] = false; 17 } 18 } 19 vector<string> permutation(string s) 20 { 21 sort(s.begin(), s.end()); 22 int n = s.length(); 23 vector<bool> vis(n, false); 24 vector<string> res; 25 string t = ""; 26 dfs(s, t, vis, res); 27 return res; 28 } 29 };