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.

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自己本来是用循环写的 结果陷入处理各种edge case的泥潭 

类似匹配括号的这种题 还是用stack吧 

以下是discuss中vote最高

This approach will just iterate through every char in the string (no recursion).

• If encounters '[', push current NestedInteger to stack and start a new one.
• If encounters ']', end current NestedInteger and pop a NestedInteger from stack to continue.
• If encounters ',', append a new number to curr NestedInteger, if this comma is not right after a brackets.
• Update index l and r, where l shall point to the start of a integer substring, while r shall points to the end+1 of substring.

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;}