341. 扁平化嵌套列表迭代器(N 叉树)
难度中等
给你一个嵌套的整数列表 nestedList
。每个元素要么是一个整数,要么是一个列表;该列表的元素也可能是整数或者是其他列表。请你实现一个迭代器将其扁平化,使之能够遍历这个列表中的所有整数。
实现扁平迭代器类 NestedIterator
:
NestedIterator(List<NestedInteger> nestedList)
用嵌套列表nestedList
初始化迭代器。int next()
返回嵌套列表的下一个整数。boolean hasNext()
如果仍然存在待迭代的整数,返回true
;否则,返回false
。
你的代码将会用下述伪代码检测:
initialize iterator with nestedList res = [] while iterator.hasNext() append iterator.next() to the end of res return res
如果 res
与预期的扁平化列表匹配,那么你的代码将会被判为正确。
示例 1:
输入:nestedList = [[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:nestedList = [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]
。
/** * // 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 { private: vector<int> res; vector<int>::iterator iter; void dfs(NestedInteger& nest) { if (nest.isInteger()) { res.emplace_back(nest.getInteger()); return; } for(auto& val : nest.getList()) { dfs(val); } } public: NestedIterator(vector<NestedInteger> &nestedList) { for(auto& val : nestedList) { dfs(val); } iter = res.begin(); } int next() { return *iter++; } bool hasNext() { return iter!=res.end(); } }; /** * Your NestedIterator object will be instantiated and called as such: * NestedIterator i(nestedList); * while (i.hasNext()) cout << i.next(); */