[LeetCode]Generate Parentheses, 解題報告

前言

發現做遞歸的題目的時候還是會遇到一些問題，有時候就怕轉不過彎來

題目

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:

思路

其實思路就是先放左括號，再放右括號，直到左右括號的數量均==n即可。注意：放括號的過程中，永遠不能出現當前右括號的數量大於左括號數量的情況

AC代碼

public class Solution {
    public static ArrayList generateParenthesis(int n) {
        ArrayList list = new ArrayList();
        StringBuilder str = new StringBuilder();

        if (n == 0) {
            return list;
        }

        recursive(0, 0, n, str, list);

        return list;
    }

    public static void recursive(int left, int right, int n, StringBuilder str, ArrayList list) {
        if (left < right) {
            return;
        }

        if (left == n && right == n) {
            String tmp = str.toString();
            list.add(tmp);
            return;
        }

        if (left < n) {
            StringBuilder newstr = new StringBuilder(str.toString());
            newstr.append('(');
            recursive(left + 1, right, n, newstr, list);
        }

        if (right < n) {
            StringBuilder newstr = new StringBuilder(str.toString());
            newstr.append(')');
            recursive(left, right + 1, n, newstr, list);
        }
    }
}