22. 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:
[
"((()))",
"(()())",
"(())()",
"()(())",
"()()()"
]
解析
- 这道题要生成正确形式的括号匹配的数量,其实就是卡特兰数,至于要输出所有括号的正确组合形式,可以采用递归。用两个变量l和r记录剩余左括号和右括号的数量,当且仅当左右括号数量都为0时,正常结束。当然还有一点限制,就是剩余的右括号数量比左括号多时才能添加右括号
class Solution_22 {
public:
void dfs(string str,int left,int right,int total,vector<string>& vec)
{
if (left+right==total)
{
vec.push_back(str);
}
if (left<total/2) // 不能用left<=total/2等号
{
dfs(str + '(', left + 1, right, total, vec);
}
if (left>right&&right<total/2) //左括号多余右括号
{
dfs(str + ')', left, right + 1, total, vec);
}
return;
}
vector<string> generateParenthesis(int n) {
vector<string> vec;
string str;
if (n==0)
{
return vec;
}
dfs("", 0, 0, 2 * n,vec);
return vec;
}
};
def generateParenthesisDFS(self, n):
def generate(s, left, right, ans):
if len(s) == 2*n:
ans.append(s)
if left < n:
generate(s+'(', left+1, right, ans)
if left > right:
generate(s+')', left, right+1, ans)
left, right, ans = 0, 0, []
generate('', left, right, ans)
return ans
def generateParenthesisBFS(self, n):
def valid(s, n):
bal, left = 0, 0
for c in s:
if c == '(':
bal += 1
left += 1
else:
bal -= 1
if bal < 0:
return False
return bal >= 0 and left <= n
ans = []
q = Queue.Queue()
q.put('(')
while not q.empty():
s = q.get()
if len(s) == 2*n:
ans.append(s)
continue
for i in ['(', ')']:
if valid(s+i, n):
q.put(s+i)
return ans
题目来源