【python】Leetcode每日一题-扁平化嵌套列表迭代器
【题目描述】
给你一个嵌套的整型列表。请你设计一个迭代器，使其能够遍历这个整型列表中的所有整数。
列表中的每一项或者为一个整数，或者是另一个列表。其中列表的元素也可能是整数或是其他列表。
示例1：
输入: [[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: [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]。
【分析】
AC代码：
# """
# This is the interface that allows for creating nested lists.
# You should not implement it, or speculate about its implementation
# """
#class NestedInteger:
# def isInteger(self) -> bool:
# """
# @return True if this NestedInteger holds a single integer, rather than a nested list.
# """
#
# def getInteger(self) -> int:
# """
# @return the single integer that this NestedInteger holds, if it holds a single integer
# Return None if this NestedInteger holds a nested list
# """
#
# def getList(self) -> [NestedInteger]:
# """
# @return the nested list that this NestedInteger holds, if it holds a nested list
# Return None if this NestedInteger holds a single integer
# """
global nested
nested = []
class NestedIterator:
def __init__(self, nestedList):
for x in nestedList:
if x.isInteger():
nested.append(x.getInteger())
else:
NestedIterator(x.getList())
def next(self) -> int:
return nested.pop(0)
def hasNext(self) -> bool:
if(len(nested) == 0):
return False
return True
有兄弟是用正则表达式直接写的
官方：
dfs
栈
public class NestedIterator implements Iterator<Integer> {
private Stack<NestedInteger> stack;
public NestedIterator(List<NestedInteger> nestedList) {
stack = new Stack<>();
for (int i=nestedList.size()-1;i>=0;i--) {
stack.push(nestedList.get(i));
}
}
@Override
public Integer next() {
NestedInteger temp = stack.pop();
return temp.getInteger();
}
@Override
public boolean hasNext() {
while (!stack.isEmpty()) {
NestedInteger temp = stack.peek();
if (temp.isInteger()) return true;
stack.pop();
for (int i=temp.getList().size()-1;i>=0;i--) {
stack.push(temp.getList().get(i));
}
}
return false;
}
}
