385. Mini Parser

Given a nested list of integers represented as a string, implement a parser to deserialize it.

Each element is either an integer, or a list -- whose elements may also be integers or other lists.

Note: You may assume that the string is well-formed:

String is non-empty.
String does not contain white spaces.
String contains only digits 0-9, [, - ,, ].
Example 1:

Given s = "324",

You should return a NestedInteger object which contains a single integer 324.
Example 2:

Given s = "[123,[456,[789]]]",

Return a NestedInteger object containing a nested list with 2 elements:

1. An integer containing value 123.
2. A nested list containing two elements:
   i. An integer containing value 456.
   ii. A nested list with one element:
   a. An integer containing value 789.

一刷
题解：
外层循坏字符串中的每个字符。

1. 如果是'[', 把当前的NestedInteger压栈并创建一个新的
2. 如果是']', 结束当前的NestedInteger，弹出NestedInteger作为当前的NestedInteger， 并把刚结束的NestedInteger加入其中
3. 如果是逗号，则把数字加入当前的NestedInteger中

所以主要的变量有：stack, curr, l, r

public NestedInteger deserialize(String s) {
    if (s.isEmpty())
        return null;
    if (s.charAt(0) != '[') // ERROR: special case
        return new NestedInteger(Integer.valueOf(s));
        
    Stack stack = new Stack<>();
    NestedInteger curr = null;
    int l = 0; // l shall point to the start of a number substring; 
               // r shall point to the end+1 of a number substring
    for (int r = 0; r < s.length(); r++) {
        char ch = s.charAt(r);
        if (ch == '[') {
            if (curr != null) {
                stack.push(curr);
            }
            curr = new NestedInteger();
            l = r+1;
        } else if (ch == ']') {
            String num = s.substring(l, r);
            if (!num.isEmpty())
                curr.add(new NestedInteger(Integer.valueOf(num)));
            if (!stack.isEmpty()) {
                NestedInteger pop = stack.pop();
                pop.add(curr);
                curr = pop;
            }
            l = r+1;
        } else if (ch == ',') {
            if (s.charAt(r-1) != ']') {
                String num = s.substring(l, r);
                curr.add(new NestedInteger(Integer.valueOf(num)));
            }
            l = r+1;
        }
    }
    
    return curr;
}

二刷：
首先要理解题意，所有的数字都会用[]包住,表示一个NestedInteger object。[[123],[456]], NestedInteger装有两个object.

三刷
还是不能直接做出来。还需要熟练。

/**
 * // 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 null if this NestedInteger holds a single integer
 *     public List getList();
 * }
 */
class Solution {
    public NestedInteger deserialize(String s) {
        if(s.isEmpty()) return null;
        if(s.charAt(0) != '[') return new NestedInteger(Integer.valueOf(s));
        
        Stack stack = new Stack<>();
        int l = 0, r = 0;
        NestedInteger cur = null;
        int val = 0;
        for(r=0; r