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leetcode-22:括号生成
2022-07-05 05:46:00 【菊头蝙蝠】
题目
数字 n 代表生成括号的对数,请你设计一个函数,用于能够生成所有可能的并且 有效的 括号组合。
示例 1:
输入:n = 3
输出:["((()))","(()())","(())()","()(())","()()()"]
示例 2:
输入:n = 1
输出:["()"]
解题
如果直接暴力回溯,让’('和‘)'自由组合,然后判断括号有效性,这样做,复杂度会比较高。
因此可以在直接通过一些判断条件,使得生成的括号是有效的。
方法一:回溯
通过open ,close来计数 开闭括号数量,保证close<open才会加入闭括号,以此来保证生成的括号的有效性。
class Solution {
public:
string path;
vector<string> res;
void dfs(int open,int close,int n){
if(path.size()==n*2){
res.push_back(path);
return;
}
if(open<n){
path.push_back('(');
dfs(open+1,close,n);
path.pop_back();
}
if(close<open){
path.push_back(')');
dfs(open,close+1,n);
path.pop_back();
}
}
vector<string> generateParenthesis(int n) {
dfs(0,0,n);
return res;
}
};
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