leetcode 385. Mini Parser | 385. 迷你语法分析器（Java）

﹏ヽ暗。殇╰゛Y 2022-08-31 06:56 20阅读 0赞

## 题目 ##

[https://leetcode.com/problems/mini-parser/][https_leetcode.com_problems_mini-parser]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3NpbmF0XzQyNDgzMzQx_size_16_color_FFFFFF_t_70]

## 题解 ##

只要有耐心分析清楚每一种状态就可以啦。

/** * // This is the interface that allows for creating nested lists. * // You should not implement it, or speculate about its implementation * public interface NestedInteger { * // Constructor initializes an empty nested list. * public NestedInteger(); * * // Constructor initializes a single integer. * public NestedInteger(int value); * * // @return true if this NestedInteger holds a single integer, rather than a nested list. * public boolean isInteger(); * * // @return the single integer that this NestedInteger holds, if it holds a single integer * // Return null if this NestedInteger holds a nested list * public Integer getInteger(); * * // Set this NestedInteger to hold a single integer. * public void setInteger(int value); * * // Set this NestedInteger to hold a nested list and adds a nested integer to it. * public void add(NestedInteger ni); * * // @return the nested list that this NestedInteger holds, if it holds a nested list * // Return empty list if this NestedInteger holds a single integer * public List<NestedInteger> getList(); * } */
    class Solution { 
        public NestedInteger deserialize(String s) { 
            if (s.charAt(0) != '[') return new NestedInteger(Integer.parseInt(s));
            Stack<NestedInteger> stack = new Stack<>();
            NestedInteger cur = new NestedInteger();
            int num = 0;
            int sn = 1; // sign=1,-1
            boolean pending = false; // 未清算标记，避免符号之间重复清算
            for (int i = 0; i < s.length(); i++) { 
                if (s.charAt(i) == ',') { 
                    // 清算上一数字
                    if (pending) { 
                        cur.add(new NestedInteger(num * sn));
                        num = 0;
                        sn = 1;
                        pending = false;
                    }
                } else if (s.charAt(i) == '[') { 
                    // 清算上一数字
                    if (pending) { 
                        cur.add(new NestedInteger(num * sn));
                        num = 0;
                        sn = 1;
                        pending = false;
                    }
                    // 当前层NI存入栈中
                    stack.push(cur);
                    // 新开一层NI
                    cur = new NestedInteger();
                } else if (s.charAt(i) == ']') { 
                    // 清算上一数字
                    if (pending) { 
                        cur.add(new NestedInteger(num * sn));
                        num = 0;
                        sn = 1;
                        pending = false;
                    }
                    // 当前层NI结束，把当前NI加入到上一层
                    NestedInteger old = stack.pop();
                    old.add(cur);
                    cur = old;
                } else if (s.charAt(i) == '-') { 
                    sn = -1;
                    pending = true;
                } else { 
                    num *= 10;
                    num += (s.charAt(i) - '0');
                    pending = true;
                }
            }
            return cur.getList().get(0); // 解开多包的一层[]
        }
    }

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3NpbmF0XzQyNDgzMzQx_size_16_color_FFFFFF_t_70 1]

[https_leetcode.com_problems_mini-parser]: https://leetcode.com/problems/mini-parser/
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3NpbmF0XzQyNDgzMzQx_size_16_color_FFFFFF_t_70]: /images/20220829/eaab8e414b7547fab9ed3d18007d23c7.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3NpbmF0XzQyNDgzMzQx_size_16_color_FFFFFF_t_70 1]: /images/20220829/a6d1ddbd0892401bb0c37c20958ba110.png