Givennpairs of parentheses, write a function togenerate all combinations of well-formed parentheses.

Input: n = 3
Output: ["((()))","(()())","(())()","()(())","()()()"]

方法一：

遞迴+遍歷。

申請2n長度的字元陣列

每位分別賦值為左右括號，當長度為2n時判斷是否合法

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，新增左括號的規則是左括號數量小於n。新增右括號的規則是右括號小於左括號。

當然方法一也可以增加新增括號規則。

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);
        }
    }

參考連結：

https://leetcode.com/problems/generate-parentheses/

https://leetcode-cn.com/problems/generate-parentheses/