[LeetCode]Generate Parentheses, 解题报告

前言

发现做递归的题目的时候还是会遇到一些问题，有时候就怕转不过弯来

题目

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

其实思路就是先放左括号，再放右括号，直到左右括号的数量均==n即可。注意：放括号的过程中，永远不能出现当前右括号的数量大于左括号数量的情况

AC代码

public class Solution {
    public static ArrayList<String> generateParenthesis(int n) {
        ArrayList<String> list = new ArrayList<String>();
        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<String> 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);
        }
    }
}