给定一个嵌套的整型列表。设计一个迭代器,使其能够遍历这个整型列表中的所有整数。
列表中的项或者为一个整数,或者是另一个列表。
示例 1:
输入: [[1,1],2,[1,1]]
输出: [1,1,2,1,1]
解释: 通过重复调用 next 直到 hasNext 返回false,next 返回的元素的顺序应该是: [1,1,2,1,1]。
示例 2:
输入: [1,[4,[6]]]
输出: [1,4,6]
解释: 通过重复调用 next 直到 hasNext 返回false,next 返回的元素的顺序应该是: [1,4,6]。
思路:这道题其实就是递归把每个嵌套的值放到一个数组中即可,但是这道题要求用栈来做,那么我们就倒着放到stack中,最后stack的栈顶元素就是数组对应的第一个元素。
参考代码:
/**
* // 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 {
public:
NestedIterator(vector<NestedInteger> &nestedList) {
ini(nestedList, vec);
}
void ini(vector<NestedInteger> nestedList,stack<int> &vec) {
for (int i = nestedList.size()-1; i >=0 ; i--) {
if (nestedList[i].isInteger()) vec.push(nestedList[i].getInteger());
else {
ini(nestedList[i].getList(),vec);
}
}
}
int next() {
int tmp = vec.top();
vec.pop();
return tmp;
}
bool hasNext() {
if (vec.empty()) return false;
return true;
}
private:
stack<int> vec;
};
/**
* Your NestedIterator object will be instantiated and called as such:
* NestedIterator i(nestedList);
* while (i.hasNext()) cout << i.next();
*/