python-剑指offer11-15
2021-01-08 11:31
标签：stack odi val app struct res 数位 使用 div 11、 输入一个整数数组，实现一个函数来调整该数组中数字的顺序，使得所有的奇数位于数组的前半部分，所有的偶数位于数组的后半部分，并保证奇数和奇数，偶数和偶数之间的相对位置不变。 12、链表 输入一个链表，输出该链表中倒数第k个结点。 解题思路：双指针。 另一种思路：使用栈 13、链表 输入一个链表，反转链表后，输出新链表的表头 14、链表 输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。 15、树 输入两棵二叉树A，B，判断B是不是A的子结构。（ps：我们约定空树不是任意一个树的子结构） python-剑指offer11-15 标签：stack odi val app struct res 数位 使用 div 原文地址：https://www.cnblogs.com/xiximayou/p/12966433.htmlclass Solution:
def reOrderArray(self, array):
# write code here
for i in range(len(array)):
for j in range(len(array)-1,i,-1):
if array[j] % 2 == 1 and array[j - 1] % 2 == 0:
array[j],array[j-1] = array[j-1],array[j]
return array
# -*- coding:utf-8 -*-
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def FindKthToTail(self, head, k):
# write code here
if head == None or k == 0:
return None
p=head
l=0
while p:
l+=1
p=p.next
if lk:
return None
p1=head
p2=head
while k!=0:
k-=1
p2=p2.next
while p2:
p2=p2.next
p1=p1.next
return p1
# -*- coding:utf-8 -*-
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def FindKthToTail(self, head, k):
# write code here
if (head == None or k == 0):
return None
count = 0
stack = []
while head != None:
stack.append(head)
head = head.next
count += 1
if (count k):
return None
for i in range(k):
res = stack.pop()
return res
# -*- coding:utf-8 -*-
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
# 返回ListNode
def ReverseList(self, pHead):
# write code here
if not pHead or pHead.next == None:
return pHead
p1=None
p2=pHead
while p2:
cur=p2.next
p2.next=p1
p1=p2
p2=cur
return p1
# -*- coding:utf-8 -*-
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
# 返回合并后列表
def Merge(self, pHead1, pHead2):
# write code here
if not (pHead1 or pHead2):
return None
elif not pHead1:
return pHead2
elif not pHead2:
return pHead1
c1 = pHead1
c2 = pHead2
head = ListNode(0)
c3 = head
while c1 and c2:
if c1.val c2.val:
c3.next = c1
c3 = c1
c1 = c1.next
else:
c3.next = c2
c3 = c2
c2 = c2.next
if c1:
c3.next = c1
if c2:
c3.next = c2
return head.next
# -*- coding:utf-8 -*-
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def HasSubtree(self, pRoot1, pRoot2):
result = False
if pRoot1 and pRoot2:
if pRoot1.val == pRoot2.val:
result = self.check_structure(pRoot1, pRoot2)
if not result:
result = self.HasSubtree(pRoot1.left, pRoot2)
if not result:
result = self.HasSubtree(pRoot1.right, pRoot2)
return result
def check_structure(self, root1, root2):
if not root2:
return True
if not root1:
return False
if root1.val != root2.val:
return False
left_check = self.check_structure(root1.left, root2.left)
right_check = self.check_structure(root1.right, root2.right)
return left_check and right_check
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