LeetCode 341 Flatten Nested List Iterator

LeetCode

You are given a nested list of integers nestedList. Each element is either an integer or a list whose elements may also be integers or other lists. Implement an iterator to flatten it.

Implement the NestedIterator class:

NestedIterator(List<NestedInteger> nestedList) Initializes the iterator with the nested list nestedList. int next() Returns the next integer in the nested list.
boolean hasNext() Returns true if there are still some integers in the nested list and false otherwise.

Your code will be tested with the following pseudocode:

initialize iterator with nestedList
res = []
while iterator.hasNext()
    append iterator.next() to the end of res
return res

If res matches the expected flattened list, then your code will be judged as correct.

Example 1:

Input: nestedList = [[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: nestedList = [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].

Constraints: 1 <= nestedList.length <= 500
* The values of the integers in the nested list is in the range [-10<sup>6</sup>, 10<sup>6</sup>].

题目大意：

实现Nested List的Iterator。Nested List是NestedInteger的数组，NestedInteger可以是int，也可以是Nested List

解题思路：

有点似括号题但区别是从前往后读取。用Queue或Stack都可以，插入或删除其一需要反序，删除nestedinteger(=list)后要将它里面所有nestedinteger加入。Queue要从队首加入和删除，Stack要从栈顶加入和删除，用Stack比较方便。

Nested List题目：
LeetCode 341 Flatten Nested List Iterator Iterator - Stack
LeetCode 339 Nested List Weight Sum - BFS
LeetCode 364 Nested List Weight Sum II - BFS

解题步骤：

N/A

注意事项：

1. 用Stack，逆序将nestedList中nestedinteger加入到stack，直到栈顶元素为int，hasNext才算结束

Python代码：

class NestedIterator(TestCases):

    def __init__(self, nestedList):
        self.stack = []
        for i in reversed(range(len(nestedList))):
            self.stack.append(nestedList[i])

    def next(self) -> int:
        if self.hasNext():
            return self.stack.pop().getInteger()
        return None

    def hasNext(self) -> bool:
        while self.stack and not self.stack[-1].isInteger():
            nestedList = self.stack.pop().getList()
            for i in reversed(range(len(nestedList))):
                self.stack.append(nestedList[i])
        return True if self.stack else False

算法分析：

next操作时间复杂度为O(V/N)或O(1)，空间复杂度O(L + N), N为所有数，V为nested list数，O(N + V)/N. O(1)如果不存在nested list