LeetCode 341 Flatten Nested List Iterator

题目

Given a nested list of integers, implement an iterator to flatten it.

Each element is either an integer, or a list – whose elements may also be integers or other lists.

Example 1:
Given the list [[1,1],2,[1,1]],

By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1,1,2,1,1].

Example 2:
Given the list [1,[4,[6]]],

By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1,4,6].

解法

/** * // This is the interface that allows for creating nested lists. * // You should not implement it, or speculate about its implementation * class NestedInteger { *   public: *     // 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; * *     // 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 NestedIterator {    vector<int>flatten_list;      int cur;    void flatten(NestedInteger i)      {          if (i.isInteger())          {              flatten_list.push_back(i.getInteger());              return;          }          vector<NestedInteger>temp = i.getList();         for (int j = 0; j < temp.size(); j++)            flatten(temp[j]);    }public:    NestedIterator(vector<NestedInteger> &nestedList) {        for (int i = 0; i < nestedList.size();i++)             flatten(nestedList[i]);        cur = 0;    }    int next() {        return flatten_list[cur++];    }    bool hasNext() {        return cur < flatten_list.size();      }};/** * Your NestedIterator object will be instantiated and called as such: * NestedIterator i(nestedList); * while (i.hasNext()) cout << i.next(); */