劍指offer刷題記錄

遞迴法：鍊錶的後續遍歷，並用self.k來記錄倒數節點的位置，找到了就返回找到的節點，否則返回none

# -*- coding:utf-8 -*-

# class listnode:
# def __init__(self, x):
# self.val = x
# self.next = none
class solution:
def __init__(self):
self.k = 0
def findkthtotail(self, head, k):
# write code here
self.k = k
def recursion(head):
if head == none:
return none
res = recursion(head.next)
if res != none:
return res
self.k = self.k - 1
if self.k == 0:
return head
return none
return recursion(head)

# -*- 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 k == 0:
return none
stack = 
while head:
head = head.next
index = 0 
while stack and index < k-1:
stack.pop()
index += 1
return stack[-1] if len(stack) > 0 else none

# -*- 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:
return none
# newhead
if phead.next == none:
return phead
# newhead
newhead = self.reverselist(phead.next)
phead.next.next = phead
phead.next = none
return newhead

# -*- coding:utf-8 -*-

# class listnode:
# def __init__(self, x):
# self.val = x
# self.next = none
class solution:
# 返回listnode
def reverselist(self, phead):
# 處理特殊情況
if phead == none:
return none
# 初始化第乙個要被翻轉的節點和其前驅
cur = phead
pre = none
# 開始迭代翻轉，並更新cur和pre
while cur:
# 先記錄下下乙個將要被翻轉的node
nxt = cur.next
cur.next = pre
pre = cur
cur = nxt
return pre

# -*- coding:utf-8 -*-

class listnode:
def __init__(self, x):
self.val = x
self.next = none
class solution:
# 返回合併後列表
def merge(self, phead1, phead2):
# 新建乙個空的head
head = listnode(0)
p = head
p1 = phead1
p2 = phead2
while p1 and p2:
if p1.val > p2.val:
# 連線
p.next = p2
# 更新p
p = p2
# 移動p2
p2 = p2.next
else:
p.next = p1
p = p1
p1 = p1.next
# 將剩餘的那邊的後續節點接到p的後面
if p1 or p2:
if p1:
p.next = p1
else:
p.next = p2
return head.next

# -*- coding:utf-8 -*-

# class listnode:
# def __init__(self, x):
# self.val = x
# self.next = none
class solution:
# 返回合併後列表
def merge(self, phead1, phead2):
# 只要有一邊節點為空了就終止遞迴, 返回結果
if not phead2:
return phead1
elif not phead1:
return phead2
if phead1.val > phead2.val:
phead2.next = self.merge(phead1, phead2.next)
return phead2
else:
phead1.next = self.merge(phead1.next, phead2)
return phead1

正確的解答在這：

# -*- 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):
# 注意這個函式不能採用求相同樹
def recur(a, b):
# b是空的話可以直接返回true (b是a的子樹)
if not b: return true
if not a or a.val != b.val: return false
return recur(a.left, b.left) and recur(a.right, b.right)
return bool(proot1 and proot2) and (recur(proot1, proot2) or self.hassubtree(proot1.left, proot2) or self.hassubtree(proot1.right, proot2))

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