LeetCode—385. Mini Parser

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>