【leetcode】341. Flatten Nested List Iterator
题目如下:
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]
.
解题思路:题目不难,递归解析即可。
代码如下:
# """ # This is the interface that allows for creating nested lists. # You should not implement it, or speculate about its implementation # """ #class NestedInteger(object): # def isInteger(self): # """ # @return True if this NestedInteger holds a single integer, rather than a nested list. # :rtype bool # """ # # def getInteger(self): # """ # @return the single integer that this NestedInteger holds, if it holds a single integer # Return None if this NestedInteger holds a nested list # :rtype int # """ # # def getList(self): # """ # @return the nested list that this NestedInteger holds, if it holds a nested list # Return None if this NestedInteger holds a single integer # :rtype List[NestedInteger] # """ class NestedIterator(object): def __init__(self, nestedList): """ Initialize your data structure here. :type nestedList: List[NestedInteger] """ self.val = [] def recursive(nestedList): for item in nestedList: if item.isInteger(): self.val.append(item.getInteger()) else: recursive(item.getList()) recursive(nestedList) def next(self): """ :rtype: int """ return self.val.pop(0) def hasNext(self): """ :rtype: bool """ return 0 != len(self.val) # Your NestedIterator object will be instantiated and called as such: # i, v = NestedIterator(nestedList), [] # while i.hasNext(): v.append(i.next())