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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: [ "((()))", "(()())", "(())()", "()(())", "()()()" ]
public class Solution { public List<String> generateParenthesis(int n) { ArrayList<String> result = new ArrayList<String>(); dfs(result, "", n, n); return result; } /* left and right represents the remaining number of ( and ) that need to be added. When left > right, there are more ")" placed than "(". Such cases are wrong and the method stops. */ public void dfs(ArrayList<String> result, String s, int left, int right){ if(left > right) //因为从left开始减一,所以left不可能小于right return; if(left==0&&right==0){ result.add(s); return; } if(left>0){ dfs(result, s+"(", left-1, right); // } if(right>0){ dfs(result, s+")", left, right-1); } } }
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