47-Generate Parentheses

1. Generate Parentheses My Submissions QuestionEditorial Solution
   Total Accepted: 86957 Total Submissions: 234754 Difficulty: Medium
   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:

“((()))”, “(()())”, “(())()”, “()(())”, “()()()”

思路：

$$f(1)="()"$$ $$f(2)="("+f(1)+")"||"()"+f(1)||f(1)+"()"$$ $$...$$ $$f(n)="( "+f(n-1)+")"||f(1)f(n-1)||f(2)f(n-2)||...||f(n-1)f(1)$$
注意，这其中计算时每次可能都会有重复，所以自底向上迭代时，在底层依次去重，可以节省大量内存空间，内层循环次数也会减少。

class Solution {
public:
    vector<string> generateParenthesis(int n) {
        vector<vector<string>> res(n+1);
        res[1].push_back("()");
        for(int i=2;i<=n;++i){    //有i组括号的时候，自底向上迭代
            vector<string> vs;
             for(int j=1;j<i;++j){    //f(j)f(i-j)的结果组合起来
                  for(int p=0;p<res[j].size();++p){
                      for(int q=0;q<res[i-j].size();++q){
                          vs.push_back(res[j][p]+res[i-j][q]);
                      }
                      if(j==i-1)vs.push_back("("+res[j][p]+")");
                  }
            }
            sort(vs.begin(),vs.end());
            vs.erase(unique(vs.begin(),vs.end()),vs.end());
            res[i] = vs;
        }
        sort(res[n].begin(),res[n].end());
        res[n].erase(unique(res[n].begin(),res[n].end()),res[n].end());
        return res[n];
    }
};