Generate Parentheses Total Accepted: 10265 Total Submissions: 33530 My Submissions
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:
"((()))", "(()())", "(())()", "()(())", "()()()"
算法應選取n個括號對，確保左括號在右括號之前，需要進行去重。要考慮時間複雜度。遞歸是解決子問題的。
class Solution {
public:
void parenthese(int p,int q,vector<string> &str,string item){
if(p==0&&q==0) {
str.push_back(item);
}else{
if(p>0){
parenthese(p-1,q,str,item+'(');
}
if(q>0&&q>p){
parenthese(p,q-1,str,item+')');遞歸帶入如此出的q-1,都不是賦值操作
}
}
}
vector<string> generateParenthesis(int n) {
vector<string> vec;
if(n<=0) return vec;
string ans;
parenthese(n,n,vec,ans);
return vec;
}
};
注意限定條件。