Generate Parentheses
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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:
((())), (()()), (())(), ()(()), ()()()
分析：
生成合法的括号串。
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递归：
每次先判断当前串中的左括号数目是否大于等于右括号数目，如果成立，那么向当前子串中添加左括号或者右括号。
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Python代码：
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class Solution(object):
def generateParenthesis(self, n):
:type n: int
:rtype: List[str]
res = []
self.dfs('', n, res)
return res
def dfs(self, cur_s, n, res):
if len(cur_s) == 2 * n:
res.append(cur_s)
return
l_n, r_n = cur_s.count('('), cur_s.count(')')
if l_n >= r_n:
if l_n < n:
self.dfs(cur_s + '(', n, res)
if r_n < n:
self.dfs(cur_s + ')', n, res)
对应的C++代码：
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class Solution {
public:
vector
generateParenthesis(int n) {
vector
res; dfs(, 0, 0, n, res); return res; } void dfs(string cur_s, int l, int r, int n, vector
& res){ if(cur_s.length() == 2 * n){ res.push_back(cur_s); return; } if(l >= r){ if(l < n){ dfs(cur_s + '(', l+1, r, n, res); } if(r < n){ dfs(cur_s + ')', l, r+1, n, res); } } } };
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别人家的代码：
Discuss中看到的动态规划：
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To generate all n-pair parentheses, we can do the following:
- Generate one pair: ()
-
Generate 0 pair inside, n - 1 afterward: () (...)...
Generate 1 pair inside, n - 2 afterward: (()) (...)...
...
Generate n - 1 pair inside, 0 afterward: ((...))
I bet you see the overlapping subproblems here. Here is the code:
(you could see in the code that x represents one j-pair solution and y represents one (i - j - 1) pair solution, and we are taking into account all possible of combinations of them)
class Solution(object):
def generateParenthesis(self, n):
:type n: int
:rtype: List[str]
dp = [[] for i in range(n + 1)]
dp[0].append('')
for i in range(n + 1):
for j in range(i):
dp[i] += ['(' + x + ')' + y for x in dp[j] for y in dp[i - j - 1]]
return dp[n] ?