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].
解题思路：借助一个栈把所有的元素先倒序存进去，每出来一个看是否是整数，如果不是依然倒序存进去，否则直接把这个数返回即可。
/** * // 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 &getList() const; * }; */class NestedIterator {public: NestedIterator(vector &nestedList) { for(int i = nestedList.size() - 1; i >= 0; i--) { s.push(nestedList[i]); } } int next() { NestedInteger top = s.top(); s.pop(); return top.getInteger(); } bool hasNext() { while(!s.empty()) { NestedInteger top = s.top(); if(top.isInteger())return true; s.pop(); for(int i = top.getList().size() - 1; i >= 0; i--) { s.push(top.getList()[i]); } } return false; }private: stack s;};/** * Your NestedIterator object will be instantiated and called as such: * NestedIterator i(nestedList); * while (i.hasNext()) cout << i.next(); */