Day15 括号生成-CFANZ编程社区

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

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

示例1：

示例2：

提示：

Java解法

package sj.shimmer.algorithm;

import java.util.ArrayList;
import java.util.List;

/**
 * Created by SJ on 2021/2/8.
 */

class D15 {
    public static void main(String[] args) {
        System.out.println(generateParenthesis(4));
    }
    public static List<String> generateParenthesis(int n) {
        List<String> strings = new ArrayList<>();
        if (n == 0) {
            strings.add("()");
            return strings;
        }
        char[] chars = new char[n * 2];
        get(strings, chars, 0);
        System.out.println(strings);
        strings.clear();
        backtrack(strings,chars,0,0);
        System.out.println(strings);
        return strings;
    }

    private static void get(List<String> strings, char[] chars, int index) {
        if (index >= chars.length) {
            int balance = 0;
            for (int i = 0; i < chars.length; i++) {
                if (chars[i]=='(') {
                    balance++;
                }else {
                    balance--;
                }
                if (balance<0) {
                    break;
                }
            }
            if (balance==0) {
                strings.add(new String(chars));
            }
            return;
        }
        chars[index]='(';
        get(strings,chars,index+1);
        chars[index]=')';
        get(strings,chars,index+1);
    }
    private static void backtrack(List<String> strings, char[] chars, int index,int leftNum) {
        if (index >= chars.length) {
            int balance = 0;
            for (int i = 0; i < chars.length; i++) {
                if (chars[i]=='(') {
                    balance++;
                }else {
                    balance--;
                }
                if (balance<0) {
                    break;
                }
            }
            if (balance==0) {
                strings.add(new String(chars));
            }
            return;
        }
        chars[index]='(';
        backtrack(strings,chars,index+1,leftNum+1);

        if (leftNum>index/2) {
            chars[index]=')';
            backtrack(strings,chars,index+1,leftNum);
        }
    }
}