Given n pairs of parentheses, write a function to generate all
combinations of well-formed parentheses.codeFor example, given n = 3, a solution set is:orm
"((()))", "(()())", "(())()", "()(())", "()()()"递归
思路:向string 中插入( 和 ),每插入一个就减1。 那么如何保证这个combination 是正确的呢?string
插入数量不超过nit
能够插入 ) 的前提是 ( 的数量大于 )io
因此就获得了递归的两个条件。function
public class Solution { public List<String> generateParenthesis(int n) { List<String> result = new ArrayList<String>(); String paren = ""; helper(result,paren, n, n); return result; } private void helper(List<String> result, String paren,int left, int right) { if (left == 0 && right == 0) { result.add(paren); return; } if (left > 0) { helper(result, paren + "(", left - 1, right); } if (right > 0 && left < right) { helper(result, paren + ")", left, right - 1); } } }