LeetCode 22. Generate Parentheses

原题链接在这里：https://leetcode.com/problems/generate-parentheses/

题目：

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:

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

题解：

DFS state needs current ( count, current ) count, current string and res.

When l==n && r==n, add cur to res and return.

If l>n or r>n, return. 

If r>l, it means there current has more ) than (, which is illegal, return.

否则先加左括号，再加右括号.

Time Complexity: exponential.

Space: O(n) 一共用了O(n)层stack.

AC Java:

class Solution {
    public List<String> generateParenthesis(int n) {
        List<String> res = new ArrayList<String>();
        if(n < 1){
            return res;
        }
        
        dfs(n, 0, 0, "", res);
        return res;
    }
    
    private void dfs(int n, int l, int r, String cur, List<String> res){
        if(l==n && r==n){
            res.add(cur);
            return;
        }
        
        if(l>n || r>n || r>l){
            return;
        }
    
        dfs(n, l+1, r, cur+'(', res);
        dfs(n, l, r+1, cur+')', res);
    }
}

AC Python:

class Solution:
    def generateParenthesis(self, n: int) -> List[str]:
        res = []
        self.dfs(n, 0, 0, "", res)
        return res
    
    def dfs(self, n: int, l: int, r: int, item: str, res: List[str]) -> None:
        if l == n and r == n:
            res.append(item)
            return
        
        if l > n or r > n or r > l:
            return
        
        self.dfs(n, l + 1, r, item + "(", res)
        self.dfs(n, l, r + 1, item + ")", res)

与Unique Binary Search Trees II类似。