python(2)，,class Solu

python(2)，,class Solu

class Solution:    def reversePrint(self, head: ListNode) -> List[int]:        newList=[]        while head:            newList.append(head.val)            head=head.next        return newList[::-1]

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class Solution:    def reversePrint(self, head: ListNode) -> List[int]:        newList=[]        if head != None:            newList.append(head.val)            head = head.next            while head !=None:                newList.append(head.val)                head=head.next            newList.reverse()        return newList

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利用递归：先走至链表末端，回溯时依次将节点值加入列表 ，这样就可以实现链表值的倒序输出。

递推阶段： 每次传入 head.next ，以 head == None（即走过链表尾部节点）为递归终止条件，此时返回空列表 [] 。回溯阶段： 利用 Python 语言特性，递归回溯时每次返回 当前 list + 当前节点值 [head.val] ，即可实现节点的倒序输出。作者：jyd链接：https://leetcode-cn.com/problems/cong-wei-dao-tou-da-yin-lian-biao-lcof/solution/mian-shi-ti-06-cong-wei-dao-tou-da-yin-lian-biao-d/来源：力扣（LeetCode）著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。class Solution:    def reversePrint(self, head: ListNode) -> List[int]:        return self.reversePrint(head.next) + [head.val] if head else []

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class CQueue:    def __init__(self):        self.stack1=[]        self.stack2=[]    def appendTail(self, value: int) -> None:        self.stack1.append(value)    def deleteHead(self) -> int:        if self.stack2:            return self.stack2.pop()        elif not self.stack1:            return -1        else:            while self.stack1:                self.stack2.append(self.stack1.pop())            return self.stack2.pop()

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使用二分查找的方式：

class Solution:    def minArray(self, numbers: [int]) -> int:        i, j = 0, len(numbers) - 1        while i < j:            m = (i + j) // 2            if numbers[m] > numbers[j]: i = m + 1            elif numbers[m] < numbers[j]: j = m            else: j -= 1        return numbers[i]

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class Solution:    def minArray(self, numbers: [int]) -> int:        return (min(numbers))        

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python(2)