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:
[ "((()))", "(()())", "(())()", "()(())", "()()()" ]
C++:
1、深搜
class Solution {
public:
void dfs(int level, int l, int r, vector<string> &solution, string s, int n){
if(2*n == level) solution.push_back(s);
if(l <= n-1)
dfs(level+1, l+1, r, solution, s+'(', n);
if(l >= r+1&&r <= n-1)
dfs(level+1, l, r+1, solution, s+')', n);
}
vector<string> generateParenthesis(int n) {
vector<string> solution;
dfs(0, 0, 0, solution, "", n);
return solution;
}
};
2、迭代
class Solution {
public:
vector<string> generateParenthesis(int n) {
vector<string> re;
if(n==0) return re;
re.push_back("");
for(int i = 0; i < 2*n; i++){
vector<string> temp;
for(int i = 0; i < re.size(); i++){
int l = 0, r = 0;
for(char x:re[i]){
if(x == '(')
l++;
if(x == ')')
r++;
}
if(l < n)
temp.push_back(re[i]+'(');
if(r+1<=l)
temp.push_back(re[i]+')');
}
re = temp;
}
return re;
}
};
3、宽搜
class Solution {
public:
vector<string> generateParenthesis(int n) {
queue<string>re;
re.push("");
int level=0;
while (level++!=2*n){
int length=re.size();
for (int i=0; i<length; ++i){
string temp=re.front();
re.pop();
int l = 0, r = 0;
for (auto y : temp){
if (y == '(') ++l;
if (y == ')') ++r;
}
if (l<n) re.push(temp + '(');
if (r<l) re.push(temp + ')');
}
}
vector<string>ans;
while(!re.empty()){
ans.push_back(re.front());
re.pop();
}
return ans;
}
};