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:
"((()))", "(()())", "(())()", "()(())", "()()()"
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class
Solution {public: vector<string> generateParenthesis(int
n) { vector<string> re; par(re,"",n,0 , 0); return
re; } void
par(vector<string> &re,string now , int
n ,int
left ,int
right) { if(left == n && right == n) { re.push_back(now); return; } if(left > right) { if(left < n) { par(re,now + ‘(‘
, n ,left+1 ,right); par(re,now + ‘)‘
, n ,left ,right+1); } else
if(left == n) { for(int
i = right ;i < n ; i++) { now+=‘)‘; } re.push_back(now); return; } } else
if(left == right) { par(re,now + ‘(‘
, n ,left+1 ,right); } }}; |
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原文：http://www.cnblogs.com/pengyu2003/p/3587690.html