标签：class ems append nbsp 回溯 div 方法 str tostring
Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
Input: n = 3
Output: ["((()))","(()())","(())()","()(())","()()()"]
方法一：
递归+遍历。
申请2n长度的字符数组
每位分别赋值为左右括号，当长度为2n时判断是否合法
public List<String> generateParenthesis(int n) { List<String> ans = new ArrayList<>(); generate(ans, new char[2 * n ], 0); return ans; } public void generate(List<String> ans, char[] cur, int n){ if( n == cur.length) { if(valid(cur)){ ans.add(new String(cur)); } return; }else{ cur[n] = ‘(‘; generate(ans, cur, n + 1); cur[n] = ‘)‘; generate(ans , cur, n + 1); } } public boolean valid(char []cur){ int balance = 0; for(char c: cur){ if(c == ‘(‘){ balance += 1; }else{ balance -= 1; } if(balance < 0) return false; } return balance == 0; }
方法二：
回溯法，添加左括号的规则是左括号数量小于n。添加右括号的规则是右括号小于左括号。
当然方法一也可以增加添加括号规则。
public List<String> generateParenthesis(int n) { List<String> ans = new ArrayList<>(); backtrack(ans, new StringBuilder(), 0 ,0 , n); return ans; } public void backtrack(List<String> ans, StringBuilder cur, int left, int right, int max){ if(cur.length() == 2 * max){ ans.add(cur.toString()); return; } if (left < max){ cur.append(‘(‘); backtrack(ans, cur, left + 1, right, max); cur.deleteCharAt(cur.length() - 1); } if( right < left ){ cur.append(‘)‘); backtrack(ans, cur, left, right + 1, max); cur.deleteCharAt(cur.length() - 1); } }
参考链接：
https://leetcode.com/problems/generate-parentheses/
https://leetcode-cn.com/problems/generate-parentheses/
标签：class ems append nbsp 回溯 div 方法 str tostring
原文地址：https://www.cnblogs.com/diameter/p/14092275.html
