Leetcode 22. Generate Parentheses - 生成指定数量的有效圆括号，比如输入2，输出()()、(())

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Given n pairs of parentheses, write a function to

generate all combinations of well-formed parentheses

.

Example 1:

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

Example 2:

Input: n = 1
Output: ["()"]

Constraints:

• 1 <= n <= 8

leetcode.com/problems/ge…

看了给定的Solution，方法1、2、3，真是难懂，全是递归.....

自己琢磨出的求解方法，非递归，容易理解很多！

我的方法（本文的方法1）: 先设置n=1，保存好()；从n=2开始，依赖上次生成的括号字符串，遍历每个括号字符串和其中的每个括号字符，生成本次的括号字符串，再把本次的结果用于计算下次的，如此循环。

字符串拼接：用StringBuilder比"String+A+B+C"的运行时间有所缩短，效率有所提高。

package com.string;
 
import java.util.*;
 
/**
 * @Author you guess
 * @Date 2022/4/10 21:41
 * @Version 1.0
 * @Desc 生成指定数量的有效圆括号 2->()()、(())
 */
public class Leetcode_22_Generate_Parentheses {
 
    /**
     * 方法1：优化后
     * Runtime: 24 ms, faster than 5.08% of Java online submissions for Generate Parentheses.
     * Memory Usage: 45.4 MB, less than 10.08% of Java online submissions for Generate Parentheses.
     *
     * @param n
     * @return
     */
    public List<String> generateParenthesis(int n) {
        List<String> list1 = new ArrayList<>();//存储总的结果
        list1.add("()");
        if (n == 1) {
            return list1;
        }
 
        //n=2 3 4 ...
        for (int i = 2; i <= n; i++) {
            Set<String> set = new HashSet<>();//存储当次的结果
            for (int j = 0; j < list1.size(); j++) {
                String s = list1.get(j);
                // set.add("()" + s);//左，多余的
                // set.add(s + "()");//右，多余的
                set.add("(" + s + ")");//外
                set.addAll(this.addParenthesis(s));
            }
            list1 = new ArrayList<>(set);
        }
        return list1;
    }
 
    /**
     * 优化后
     */
    public Set<String> addParenthesis(String s) {
        Set<String> set = new HashSet<>();
        Stack<Integer> stack = new Stack<>();
        for (int i = 0; i < s.length(); i++) {
            StringBuilder sb = new StringBuilder();
            char c = s.charAt(i);
            if (c == '(') {
                //()()(()) -> (()())(()),这个比较容易忘记！！
                if (stack.isEmpty()) {
                    sb.append("(").append(s, 0, i).append(")").append(s.substring(i));
                    set.add(sb.toString());
                    //set.add("(" + s.substring(0, i) + ")" + s.substring(i));
                }
                stack.push(i);
            } else {
                int left = stack.pop();
                int right = i;
                //在里面加一层(),比如([待加入])() -> ([()])()
                sb.append(s, 0, left + 1).append("(").append(s, left + 1, right).append(")").append(s.substring(right));
                set.add(sb.toString());
                //set.add(s.substring(0, left + 1) + "(" + s.substring(left + 1, right) + ")" + s.substring(right));
            }
        }
        return set;
    }
 
 
    class Entry {
        Character c;
        Integer index;
 
        public Entry(Character c, Integer index) {
            this.c = c;
            this.index = index;
        }
    }
 
    /**
     * 这个也正确，不过不必使用Entry类
     * 未优化
     *
     * @param s
     * @return
     */
    public Set<String> addParenthesisUseEntry(String s) {
        Set<String> set = new HashSet<>();
        Stack<Entry> stack = new Stack<>();
 
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (c == '(') {
                //()()(()) -> (()())(()),这个比较容易忘记！！
                if (stack.isEmpty()) {
                    set.add("(" + s.substring(0, i) + ")" + s.substring(i));
                }
                Entry leftEntry = new Entry(s.charAt(i), i);
                stack.push(leftEntry);
            } else {
                Entry leftEntry = stack.pop();
                int left = leftEntry.index;// int left = leftEntry.getIndex() ; 同义
                int right = i;
                set.add(s.substring(0, left + 1) + "(" + s.substring(left + 1, right) + ")" + s.substring(right));
            }
        }
        return set;
    }
 
 
    public static void main(String[] args) {
 
        Leetcode_22_Generate_Parentheses main = new Leetcode_22_Generate_Parentheses();
        //System.out.println(main.generateParenthesis(3));
        System.out.println(main.generateParenthesisSolution(3));
 
    }
}