Flatten Nested List Iterator
Desicription
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:
代码语言：javascript复制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:
代码语言：javascript复制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].Solution
代码语言：javascript复制/**
* // 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:
std::vector<int> integerVector;
int index;
public:
explicit NestedIterator(const std::vector<NestedInteger>& nestedList) {
index = 1;
integerVector = std::vector<int>(1);
std::function<void(const std::vector<NestedInteger>& nestedLis)> dfs = [&](const std::vector<NestedInteger>& roots) {
for(const auto& root : roots) {
if(root.isInteger()) {
integerVector.push_back(root.getInteger());
} else {
dfs(root.getList());
}
}
};
dfs(nestedList);
}
int next() {
return integerVector[index ];
}
bool hasNext() {
return index <= integerVector.size() - 1;
}
};
/**
* Your NestedIterator object will be instantiated and called as such:
* NestedIterator i(nestedList);
* while (i.hasNext()) cout << i.next();
*/
