Leetcode-22. 括号生成

给出 n 代表生成括号的对数，请你写出一个函数，使其能够生成所有可能的并且有效的括号组合。

例如，给出 n = 3，生成结果为：

[
“((()))”,
“(()())”,
“(())()”,
“()(())”,
“()()()”
]

解法：还是用DFS，到底之后回溯，到上一步之后看还能用哪个括号，大概就是这样的思路，dfs代码写起来都是比较简洁，但是比较难想难理解

  • Java 
class Solution {
   
    public List<String> generateParenthesis(int n) {
   
        List<String> lst = new ArrayList<>();
        this.helper(lst, n, 0, 0, "");
        return lst;
    }
    public void helper(List<String> lst, int n, int leftUsed, int rightUsed, String str) {
   
        if(leftUsed==n && rightUsed==n) lst.add(str);
        if(leftUsed < n) this.helper(lst, n, leftUsed+1, rightUsed, str+"(");
        if(leftUsed > rightUsed && rightUsed < n) this.helper(lst, n, leftUsed, rightUsed+1, str+")");
    } 
}
  • Python 
class Solution:
    def generateParenthesis(self, n: int) -> List[str]:
        self.lst = []
        self.helper(n, 0, 0, "")
        return self.lst
    
    def helper(self, n, leftUsed, rightUsed, str):
        if(leftUsed==n and rightUsed==n): self.lst.append(str)
        if(leftUsed < n):
            self.helper(n, leftUsed+1, rightUsed, str+"(")
        if(rightUsed < n and leftUsed>rightUsed):
            self.helper(n, leftUsed, rightUsed+1, str+")")