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.
这道题让我们实现一个迷你解析器用来把一个字符串解析成NestInteger类，关于这个嵌套链表类的题我们之前做过三道，
Nested List Weight Sum II，Flatten Nested List Iterator，和Nested List Weight Sum。

注: Integer.valueof(String s)是将一个包装类是将一个实际值为数字的变量先转成string型再将它转成Integer型的包装类对象(相当于转成了int的对象)这样转完的对象就具有方法和属性了。

而Integer.parseInt(String s)只是将是数字的字符串转成数字，注意他返回的是int型变量不具备方法和属性

 Integer.parseInt()把String 型转换为Int型，

 Integer.valueOf()把String 型转换为Integer对象。

大概知道一点了，就是说Integer.valueOf(S)是针对包装类来说的，而Integer.parseInt(s) 是针对变量而言.

括号的题想到用栈, 并不断判断符号来操作, 生成对象和更新对象, 对象的生成在字符串里通常靠双指针 

心得: 字符串的题, 要学会考虑多种情况是用substring 还是charAt判断, 根据题意什么时候用substring

/**
 * // 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<NestedInteger> getList();
 * }
 */

　　

 

public NestedInteger deserialize(String s) {
    if (s.isEmpty())
        return null;
    if (s.charAt(0) != '[') // ERROR: special case
        return new NestedInteger(Integer.valueOf(s));
        
    Stack<NestedInteger> 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;
}

一看要求就是用栈, 遍历的时候得知道遇到什么pop, 什么push, 什么时候移动左窗口指针(窗口旨在构造类, 截取元素), 

遇到不同的符号怎么处理. 

之所以用栈是因为题目要求和栈的特点符合: 能push, 能pop