【dfs,回溯】22. 括号生成

数字 n 代表生成括号的对数，请你设计一个函数，用于能够生成所有可能的并且 有效的 括号组合。

输入：n = 3输出：[
   "((()))",
   "(()())",
   "(())()",
   "()(())",
   "()()()"
 ]

非回溯法：

放到参数中，每次新建字符串

import org.w3c.dom.html.HTMLHeadElement;

import java.util.*;

class Solution {
    public List<String> generateParenthesis(int n) {
        return dfs(n, n, "");
    }

    private List<String> dfs(int left, int right, String s) {
        List<String> res = new ArrayList<>();
        if (left == 0 && right == 0) {
            res.add(s);
        }
        if (left > 0)
            res.addAll(dfs(left - 1, right, s + "("));
        if (right > left)
            res.addAll(dfs(left, right - 1, s + ")"));
        return res;
    }

    public static void main(String[] args) {
        System.out.println(new Solution().generateParenthesis(4));
    }
}

回溯法：

import java.util.*;

class Solution {
    public List<String> generateParenthesis(int n) {
        StringBuilder sb = new StringBuilder();
        return dfs(n, n, sb);
    }

    private List<String> dfs(int left, int right, StringBuilder sb) {
        List<String> res = new ArrayList<>();
        if (left == 0 && right == 0) {
            res.add(sb.toString());
            return res;
        }
        if (left > 0) {
            res.addAll(dfs(left - 1, right, sb.append("(")));
            sb.deleteCharAt(sb.length() - 1);//回溯，删除上次加的字符
        }
        if (right > left) {
            res.addAll(dfs(left, right - 1, sb.append(")")));
            sb.deleteCharAt(sb.length() - 1);//回溯，删除上次加的字符
        }
        return res;
    }

    public static void main(String[] args) {
        System.out.println(new Solution().generateParenthesis(4));
    }
}