341. Flatten Nested List Iterator
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:
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]
Example 2:
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]
题目链接：https://leetcode-cn.com/problems/flatten-nested-list-iterator/
解题思路
从类的调用效率来看，有2种设计：
设计一：时间均衡
构造时只做初步拆解，在调用hasNext时进一步拆。
一边拆解一边调用，运行时间较分散。
设计二：头重脚轻
在构造时完成所有拆解。
构造费时，调用快。
从拆解方法上，有2中方法：
法一：递归
将对list的拆解写成函数，遇到list则向下调用。
法二：非递归，栈
将list逆向压入栈，不断取栈顶解析。
结合设计和方法，思考如何实现代码：
设计一+法二：递归一条道走到黑，和拆分时间的思想不一致。
设计二+法一：递归和一股脑拆解完的思路不谋而合；递归中已经使用了栈来存储临时状态，没有必要再去设计栈。
代码实现
设计一+法二
注意：压栈时需要对vector里的元素逆着取。
/**
* // 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 {
stack<NestedInteger> record;
public:
NestedIterator(vector<NestedInteger> &nestedList) {
if(nestedList.empty()) return;
for(auto iter=nestedList.rbegin(); iter!=nestedList.rend(); ++iter){
record.push(*iter);
}
}
int next() {
auto tmp = record.top();
record.pop();
return tmp.getInteger();
}
bool hasNext() {
while(!record.empty()){
auto tmp = record.top();
if(tmp.isInteger()){
return true;
}
record.pop();
auto tmplist = tmp.getList();
for(auto iter=tmplist.rbegin(); iter!=tmplist.rend(); ++iter){
record.push(*iter);
}
}
return false;
}
};
/**
* Your NestedIterator object will be instantiated and called as such:
* NestedIterator i(nestedList);
* while (i.hasNext()) cout << i.next();
*/
设计二+法一
/**
* // 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> record;
int idx;
public:
NestedIterator(vector<NestedInteger> &nestedList) {
if(nestedList.empty()) return;
flaten(nestedList);
idx = 0;
}
void flaten(vector<NestedInteger> &list){
for(auto iter=list.begin(); iter!=list.end(); ++iter){
if((*iter).isInteger()){
record.push_back((*iter).getInteger());
}else{
flaten((*iter).getList());
}
}
}
int next() {
return record[idx++];
}
bool hasNext() {
if(idx<record.size()) return true;
else return false;
}
};
/**
* Your NestedIterator object will be instantiated and called as such:
* NestedIterator i(nestedList);
* while (i.hasNext()) cout << i.next();
*/
来源：CSDN
作者：lemonade13
链接：https://blog.csdn.net/lemonade13/article/details/104293531