Leetcode Generate Parentheses

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:

"((()))", "(()())", "(())()", "()(())", "()()()"




Idea:

A tricky backtracking problem. Given n pairs of parentheses, there will be n '(' and n ')'. If we have already had l '(' and r ')' in the previous part string, if l==n, next time we can only append next ')' go the previous part to get a valid combination. if l<n, next time we can append one '(' to the previous string and use backtracking the consider next condition. if l<n while r<l, this time we also can add one ')', and go to next recursive function.


code:

public class Solution {    public List<String> generateParenthesis(int n) {        List<String> res = new ArrayList<String>();        StringBuffer buffer = new StringBuffer();        dfs(res, buffer, n, 0, 0);        return res;    }    private void dfs(List<String> res, StringBuffer buffer, int n, int left, int right){        if(left==n){            int size = buffer.length();            for(int i=0;i<left-right;i++){                buffer.append(')');            }            res.add(buffer.toString());            buffer.setLength(size);            return;        }        buffer.append('(');        dfs(res, buffer, n, left+1, right);        buffer.setLength(buffer.length()-1);        if(left>right){            buffer.append(')');            dfs(res, buffer, n, left, right+1);            buffer.setLength(buffer.length()-1);        }    }}