Leetcode 385. Mini Parser

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题意
这题看题看了好久，不太理解题意。觉得Leetcode的难度在于理解题目上面。

题意是根据所给的字符串，解析出一个NestedInteger的对象。
解析的方式就是最简单的自动机。
因为很类似递归处理，所以用一个栈来维护每一层的NestedInteger。

然后返回最顶层的就行了。

(Leetcode代码收录](https://github.com/Mr-Phoebe/ACM-ICPC/tree/master/OJ/Leetcode)，求粉求星星。

/** * // This is the interface that allows for creating nested lists. * // You should not implement it, or speculate about its implementation * class NestedInteger { *   public: *     // Constructor initializes an empty nested list. *     NestedInteger(); * *     // Constructor initializes a single integer. *     NestedInteger(int value); * *     // Return true if this NestedInteger holds a single integer, rather than a nested list. *     bool isInteger() const; * *     // Return the single integer that this NestedInteger holds, if it holds a single integer *     // The result is undefined if this NestedInteger holds a nested list *     int getInteger() const; * *     // Set this NestedInteger to hold a single integer. *     void setInteger(int value); * *     // Set this NestedInteger to hold a nested list and adds a nested integer to it. *     void add(const NestedInteger &ni); * *     // Return the nested list that this NestedInteger holds, if it holds a nested list *     // The result is undefined if this NestedInteger holds a single integer *     const vector<NestedInteger> &getList() const; * }; */class Solution {  public:      NestedInteger deserialize(string s) {          if(s[0] != '[') return NestedInteger(stoi(s));          stack<NestedInteger*> stk;          NestedInteger* ans = NULL;          int idx = 0;          for(int i=0; i<s.size(); i++) {              if(s[i] == '[') {                  stk.push(new NestedInteger());                  if(!ans) ans = stk.top();                  idx = i + 1;              }              else if(s[i] == ',') {                if(idx != i)                      stk.top()->add(NestedInteger(stoi(s.substr(idx, i-idx))));                  idx = i + 1;            }            else if(s[i] == ']') {                  if(idx != i)                      stk.top()->add(NestedInteger(stoi(s.substr(idx, i-idx))));                 NestedInteger* cur = stk.top();                  stk.pop();                  if(!stk.empty()) stk.top()->add(*cur);                  idx = i + 1;              }          }          return *ans;      }  };

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