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在LeetCode中生成括号

[英]Generate Parentheses in LeetCode

I'm working on solving problem Generate Parentheses on LeetCode . 我正在解决问题在LeetCode生成括号 My first thought was that the n+1 th pair of parentheses could be deduced from n th pair of parentheses. 我的第一个想法是可以从第n对括号中推导出第n+1对括号。 Saying, if we use e as the situation of n th, and I thought the n+1 th will be one of the following situation: 说,如果我们使用e作为第n个情况,而我认为第n+1个情况将是以下情况之一:

  • ( + e + ) ( + e + )
  • () + e () + e
  • e + () e + ()

And I came up with the following code. 我想出了以下代码。

public class Solution {
    public List<String> generateParenthesis(int n) {
        HashSet<String> temp = new HashSet<String>();
        HashSet<String> result = new HashSet<String>();
        List<String> rsult = new ArrayList<String>();

        if (n == 0) {
            temp.add("");
            return new ArrayList<String>(temp);
        }

        temp.add("()");

        if (n == 1) {

            return new ArrayList<String>(temp);
        }

        for (int i = 2; i <= n; i++) {
            result.removeAll(temp);
            for (String e : temp) {
                if (!result.contains("(" + e + ")")) {
                    result.add("(" + e + ")");
                }

                if (!result.contains("()" + e)) {
                    result.add("()" + e);
                }

                if (!result.contains(e + "()")) {
                    result.add(e + "()");
                }

            }

            temp.clear();
            temp.addAll(result);
        }
        rsult = new ArrayList<String>(result);
        Collections.sort(rsult);
        return rsult;
    }
}

However, when I submitted the code, I found that I still missed some cases when the n+1 is even. 但是,当我提交代码时,我发现我仍然错过了n+1为偶数的情况。 So I updated my code as below. 所以我更新了我的代码如下。

public class Solution {
    public List<String> generateParenthesis(int n) {
        HashSet<String> temp = new HashSet<String>();
        HashSet<String> result = new HashSet<String>();
        List<String> rsult = new ArrayList<String>();

        if (n == 0) {
            temp.add("");
            return new ArrayList<String>(temp);
        }

        temp.add("()");

        if (n == 1) {

            return new ArrayList<String>(temp);
        }

        for (int i = 2; i <= n; i++) {
            result.removeAll(temp);
            for (String e : temp) {
                if (!result.contains("(" + e + ")")) {
                    result.add("(" + e + ")");
                }

                if (!result.contains("()" + e)) {
                    result.add("()" + e);
                }

                if (!result.contains(e + "()")) {
                    result.add(e + "()");
                }

                if (i % 2 == 0) {
                    String dblprt = new String();
                    for(int j = 0; j< i/2;j++) {
                        dblprt = "(" + dblprt + ")";
                    }
                    dblprt = dblprt + dblprt ;
                    if (!result.contains(dblprt)) {
                        result.add(dblprt);
                    }
                }
            }

            temp.clear();
            temp.addAll(result);
        }
        rsult = new ArrayList<String>(result);
        Collections.sort(rsult);
        return rsult;
    }
}

Still, the test cases failed. 尽管如此,测试用例还是失败了。 So I'm confused. 所以我很困惑。 Why doesn't my way work? 为什么我的方式不起作用? Am I still missing some cases? 我还缺少一些案件吗?

Why doesn't my way work? 为什么我的方式不起作用? Am I still missing some cases? 我还缺少一些案件吗?

Consider the parentheses (())(()) , the only way for this to have come from your algorithm is if e = ())(() , and then ( + e + ) but in that case e is not a well-formed parentheses so e never existed and you have missed a case. 考虑括号(())(()) ,从算法中得出的唯一方法是,如果e = ())(() ,然后是( + e + )但在这种情况下, e不太适合形式的括号,因此e永远不存在,并且您错过了一个案例。

EDIT : it would seem that your second revision solves cases such as ()() , (())(()) , ((()))((())) but what about (()())(()()) , or (()())()(()) ? 编辑 :似乎您的第二个修订版解决了诸如()()(())(())((()))((()))(()())(()())还是(()())()(())

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