第100天。
今天的题目是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: 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].
挺有趣的题目，主要是要实现一个嵌套列表的迭代器，大概是三个函数：
class NestedIterator {
public:
NestedIterator(vector<NestedInteger> &nestedList) {
}
int next() {
}
bool hasNext() {
}
};
然后他也提供了NestedInteger的接口和一些说明，算是对题目的补充：
/**
* // 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;
* };
*/
/**
* Your NestedIterator object will be instantiated and called as such:
* NestedIterator i(nestedList);
* while (i.hasNext()) cout << i.next();
*/
从补充中我们可以看到，在调用next前一定会先调用hasNext,有了这个前提我们写起来会方便一点。
我的解法是，NestedIterator只保存构造函数中传入的nestedList的两个迭代器，之所以是两个，是因为要保存end迭代器，然后要实现嵌套，我们还要一个NestedIterator的指针，利用这个指针来对下一级列表的元素进行迭代：
class NestedIterator {
public:
NestedIterator(vector<NestedInteger> &nestedList) {
_it = nestedList.begin(); _end_it = nestedList.end();
_tmp_it = nullptr;
}
int next() {
if (_it->isInteger()) { int ret = _it->getInteger(); ++_it; return ret; }
return _tmp_it->next();
}
bool hasNext() {
if (_it == _end_it) return false;
if (_it->isInteger()) return true;
if (_tmp_it == nullptr) {
_tmp_it = new NestedIterator(_it->getList());
}
if (_tmp_it->hasNext()) return true;
delete _tmp_it; _tmp_it = nullptr; ++_it;
return hasNext();
}
private:
vector<NestedInteger>::iterator _it, _end_it;
NestedIterator *_tmp_it;
};
然后这里的实现虽然比较简单，简洁，但是在遇到一些特殊情况的时候会对性能造成极大的影响,比如说[[[[[1,2,3]]]]],虽然只有三个元素，但是因为有5层的嵌套，我们要有5个迭代器，每次调用next和hasNext都需要递归调用5次才能返回，这样效率就有点低了.
dicuss中的解法会比较好一点，类别DFS来做，先用stack保存所有的元素，在调用hasNext的时候，如果栈顶是列表就将其展开并压栈（倒序），然后在递归调用hasNext，直到栈顶为数字时,然后调用next就直接返回栈顶即可：
class NestedIterator {
public:
NestedIterator(vector<NestedInteger> &nestedList) {
begins.push(nestedList.begin());
ends.push(nestedList.end());
}
int next() {
hasNext();
return (begins.top()++)->getInteger();
}
bool hasNext() {
while (begins.size()) {
if (begins.top() == ends.top()) {
begins.pop();
ends.pop();
} else {
auto x = begins.top();
if (x->isInteger())
return true;
begins.top()++;
begins.push(x->getList().begin());
ends.push(x->getList().end());
}
}
return false;
}
private:
stack<vector<NestedInteger>::iterator> begins, ends;
};
update at 2020-04-03
和第二个解法有点像，但是只需要一个栈即可：
class NestedIterator {
public:
stack<vector<NestedInteger>::iterator> st;
NestedIterator(vector<NestedInteger> &nestedList) {
reversePush(nestedList);
}
void reversePush(vector<NestedInteger> &nestedList) {
auto beg = nestedList.begin();
for(int i = nestedList.size() - 1; i >= 0; --i) {
st.push(beg + i);
}
}
int next() {
auto top = st.top(); st.pop();
return top->getInteger();
}
bool hasNext() {
while(!st.empty() && !st.top()->isInteger()) {
auto top = st.top(); st.pop();
reversePush(top->getList());
}
return !st.empty();
}
};