---
# System prepended metadata

title: 22. Generate Parentheses
tags: [Top 100 Liked Questions, LeetCode]

---

---
tags: LeetCode,Top 100 Liked Questions
---
# 22. Generate Parentheses
https://leetcode.com/problems/generate-parentheses/submissions/
## 題目敘述
Given a string s containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.

An input string is valid if:

Open brackets must be closed by the same type of brackets.
Open brackets must be closed in the correct order.
## Example
Example 1:
```
Input: n = 3
Output: ["((()))","(()())","(())()","()(())","()()()"]
```
Example 2:
```
Input: n = 1
Output: ["()"]
```
## 解題想法
### 1.backtracking
base case:current length=max*2 and 左括號數量=右括號數量
且每次放進current的可以是左括號或右括號(左括號數量<max and 左括號數量>右括號數量)
```
class Solution {
public:
    void backtracking(vector<string> &res,string current,int open,int close,int max){
    //&res要記得傳位址才能改變res
        if(current.length()==max*2){/base case
            res.push_back(current);
            return;
        }
        if(open<max){//左括號數量<max
            current+="(";
            backtracking(res,current,open+1,close,max);
            current.pop_back();//要把加上的括號去掉
        }
        if(close<open){//左括號數量<max and 左括號數量>右括號數量
            current+=")";
            backtracking(res,current,open,close+1,max);
            current.pop_back();//要把加上的括號去掉
        }
    }
    vector<string> generateParenthesis(int n) {
        vector<string> res;
        backtracking(res,"",0,0,n);
        return res;
    }
    
};
```
時間複雜度為O(N^2)
