19.1.29 [LeetCode 22] Generate Parentheses

Posted 2021-02-06 TobicYAL

Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.

For example, given n = 3, a solution set is:

[
  "((()))",
  "(()())",
  "(())()",
  "()(())",
  "()()()"
]

 1 class Solution {
 2 public:
 3     vector<string>build(int toclose, int toopen) {
 4         vector<string>ans;
 5         if (toclose == 0 && toopen == 0) {
 6             ans.push_back("");
 7             return ans;
 8         }
 9         if (toclose > 0) {
10             vector<string>closeone = build(toclose - 1, toopen);
11             for (int i = 0; i < closeone.size(); i++)
12                 ans.push_back(")" + closeone[i]);
13         }
14         if (toopen > 0) {
15             vector<string>openone = build(toclose + 1, toopen - 1);
16             for (int i = 0; i < openone.size(); i++)
17                 ans.push_back("(" + openone[i]);
18         }
19         return ans;
20     }
21     vector<string> generateParenthesis(int n) {
22         return build(0, n);
23     }
24 };

View Code

一开始使用了递归方法，很慢，然后我看到还有其他解法是这样的??

 1 class Solution {
 2 public:
 3     vector<string> generateParenthesis(int n) {
 4         vector<string>ans;
 5         if (n == 0) {
 6             ans.push_back("");
 7             return ans;
 8         }
 9         set<string>_ans;
10         vector<string>tmp = generateParenthesis(n - 1);
11         for (int i = 0; i < tmp.size(); i++) {
12             string now = tmp[i];
13             _ans.insert("()" + now);
14             for (int j = 0; now[j]; j++) {
15                 if (now[j] == ‘(‘)
16                     _ans.insert(now.substr(0, j + 1) + "()" + now.substr(j + 1));
17             }
18         }
19         return vector<string>(_ans.begin(), _ans.end());
20     }
21 };

View Code

这种方法很奇妙，是基于n-1递推出n的

因为所有n解都可以删去一个完整的()获得一个n-1解，所以在这里只要在n-1解上可以加()的位置加上一个()并且排除重复解即可得到结果