Mini Parser思路：1.首先根据是否有括号判定是否为纯数字，如果是纯数字则直接转化。2其次判断list是否为空，为空直接转化。3.统计左右括号是否对应，如果对应，则遇到逗号后递归处理，无论是list还是纯数字。
GitHub地址：https://github.com/corpsepiges/leetcode
目前java版本的答案大约进度是免费的差40题，python大约是一半，其他的等以后再补充。
点此进入如果可以的话，请点一下star，谢谢。
<span style="font-size:12px;">/** * // 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 class Solution { public NestedInteger deserialize(String s) { if (s.contains("[")) { NestedInteger ans=new NestedInteger(); if (s.length()>2) { int begin=1; char[] cs=s.toCharArray(); int count=0; for (int i = 1; i < s.length()-1; i++) { if (cs[i]==','&&count==0) { ans.add(deserialize(s.substring(begin,i))); begin=i+1; } if (cs[i]=='['||cs[i]==']') { count+=(92-cs[i]); } } ans.add(deserialize(s.substring(begin,s.length()-1))); } return ans; } return new NestedInteger(Integer.valueOf(s)); } }</span>
