11/* Generate Parentheses:Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.*/
2+ // 一开始left = n,right = 0;当left还剩余左括号数大于0的时候,添加左括号,并将left – 1,right + 1;
3+ // 当right目前还需要的右括号数大于0的时候,添加右括号,并将right – 1;
4+ // 当left和right都等于0的时候,表示当前是一个符合格式条件的字符串,将该字符串cur放入result数组中,最后返回result。
25class Solution {
36public:
47 vector<string> generateParenthesis (int n) {
5- vector<string> rs;
6- string s;
7- genParenthesis (rs, s, n, n);
8- return rs;
8+ dfs (" " 0 );
9+ return result;
10+ }
11+ private:
12+ vector<string> result;
13+ void dfs (string cur, int left, int right) {
14+ if (left == 0 && right == 0 ) {
15+ result.push_back (cur);
16+ return ;
17+ }
18+ if (left > 0 ) dfs (cur + " (" 1 , right + 1 );
19+ if (right > 0 ) dfs (cur + " )" 1 );
920 }
10- void genParenthesis (vector<string> &rs, string &s, int left, int right)
11- {
12- if (left==0 )
13- {
14- rs.push_back (s);
15- rs.back ().append (right, ' )'
16- return ;
17- }
18- s.push_back (' ('
19- genParenthesis (rs, s, left-1 , right);
20- s.pop_back ();
21- if (left < right)
22- {
23- s.push_back (' )'
24- genParenthesis (rs, s, left, right-1 );
25- s.pop_back ();
26- }
27- }
2821};
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