题目链接：https://leetcode.com/problems/mini-parser/ 题目：
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.
思路：
用栈维护一个包含关系，类似于用栈维护带 '(' 的表达式。
思路不难想到，有个坑爹的细节要注意：添加一个数到list type的nestedInteger时 要将该数封装为integer type的NestedInteger，然后用add方法添加该nestedInteger，不能直接用setInteger，否则会把list type的nestedInteger定义为一个integer type，会出错。
还可以用递归来做，思路大致是：没处理完一个token，遇到 [ 递归处理 返回该层list结尾下标。好像有点麻烦。。留坑日后待填。
算法：
public NestedInteger deserialize(String s) { Stack<NestedInteger> stack = new Stack<NestedInteger>(); String tokenNum = ""; for (char c : s.toCharArray()) { switch (c) { case '['://[代表一个list stack.push(new NestedInteger()); break; case ']'://list结尾 if (!tokenNum.equals(""))//前面token为数字 stack.peek().add(new NestedInteger(Integer.parseInt(tokenNum)));//将数字加入到本层list中 NestedInteger ni = stack.pop();//本层list结束 tokenNum = ""; if (!stack.isEmpty()) {//栈内有更高层次的list stack.peek().add(ni); } else {//栈为空，遍历到字符串结尾 return ni; } break; case ',': if (!tokenNum.equals("")) //将数字加入到本层list中 stack.peek().add(new NestedInteger(Integer.parseInt(tokenNum))); tokenNum = ""; break; default: tokenNum += c; } } if (!tokenNum.equals(""))//特殊case: 如果字符串只包含数字的情况 return new NestedInteger(Integer.parseInt(tokenNum)); return null; }
