题目描述
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.
Example1:
Input: [[1,1],2,[1,1]] Output: [1,1,2,1,1] Explanation: By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1,1,2,1,1].
Example2:
Input: [1,[4,[6]]] Output: [1,4,6] Explanation: By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1,4,6].
解题思路
- 参考
- 我们在对向量遍历的时候，从后往前把对象压入栈中，那么第一个对象最后压入栈就会第一个取出来处理，
- 我们的hasNext()函数需要遍历栈，并进行处理，如果栈顶元素是整数，直接返回true，如果不是，那么移除栈顶元素，并开始遍历这个取出的list，还是从后往前压入栈，循环停止条件是栈为空，返回false，
/**
* // 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) {
for(int i = nestedList.size() - 1; i >= 0; i--){
s.push(nestedList[i]);
}
}
int next() {
NestedInteger t = s.top();
s.pop();
return t.getInteger();
}
bool hasNext() {
while(!s.empty()){
NestedInteger t = s.top();
if(t.isInteger()){
return true;
}
s.pop();
for(int i = t.getList().size() - 1; i >= 0; i--){
s.push(t.getList()[i]);
}
}
return false;
}
private:
stack<NestedInteger> s;
};
/**
* Your NestedIterator object will be instantiated and called as such:
* NestedIterator i(nestedList);
* while (i.hasNext()) cout << i.next();
*/
