Description:
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:
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].
Link:
https://leetcode.com/problems/flatten-nested-list-iterator/description/
解題方法：
NestedIterator：
將vector里面的NestedInteger反過來亞入棧。
hasNext：
如果棧頂的NestedInteger是一個單獨的數則返回true，否則從棧頂的這個List取出其中的NestedInteger，再反過來壓棧，直到棧頂的元素是單獨的數。
next：
從棧頂取出數即可。
Time Complexity：
NestedIterator：O(N)
next：O(1)
完整代碼：
class NestedIterator
{
private:
stack<NestedInteger> container;
public:
NestedIterator(vector<NestedInteger> &nestedList)
{
for(int i = nestedList.size() - 1; i >= 0; i--)
{
container.push(nestedList[i]);
}
}
int next()
{
int result = container.top().getInteger();
container.pop();
return result;
}
bool hasNext()
{
while(!container.empty())
{
NestedInteger curr = container.top();
if(curr.isInteger())
return true;
container.pop();
vector<NestedInteger> & currList = curr.getList();
for(int i = currList.size() - 1; i >= 0; i--)
{
container.push(currList[i]);
}
}
return false;
}
};
