從尾到頭列印鍊錶
class solution:
# 返回從尾部到頭部的列表值序列，例如[1,2,3]
def printlistfromtailtohead(self, listnode):
rs =
if not listnode:return rs
while listnode:
listnode = listnode.next
return rs[::-1]
# -*- coding:utf-8 -*-
class listnode:
def __init__(self, x):
self.val = x
self.next = none
class solution:
def deleteduplication(self, phead):
# write code here
if not phead:return phead
dummy = listnode(-1)
dummy.next = phead
p = dummy
while p.next:
# 原來錯誤是while dummy.next：
# dummy在while中都沒有發生改變，起不到迴圈結束的作用
q = p.next
while (q and p.next.val == q.val):
q = q.next
# 如果不存在重複，長度==1
if p.next.next == q:
p = p.next
# 如果存在重複，長度》1
else: p.next = q
return dummy.next
class solution:
def reverselist(self, phead):
# 反轉鍊錶需要三個listnode
pre = none
cur = phead
while cur:
tail = cur.next
cur.next = pre
pre = cur
cur = tail
# return cur # cur目前指的是空
return pre
class solution:
def findfirstcommonnode(self, phead1, phead2):
if not phead1:return phead1
if not phead2:return phead2
len1 = self.getlength(phead1)
len2 = self.getlength(phead2)
diff = abs(len1-len2)
p1 = phead1
p2 = phead2
if len1>len2:
while diff>0:
diff -= 1
p1= p1.next
if len10:
diff -= 1
p2 = p2.next
# print(p1.val)
# print(p2.val)
while p1 != p2:
p1 = p1.next
p2 = p2.next
return p1
def getlength(self,head):
cnt = 0
cur = head
while cur:
cnt+=1
cur = cur.next
return cnt
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):
# write code here
p1,p2 = none,none
rs = none
head = rs
while phead1 and phead2:
if phead1.val < phead2.val:
rs.next = phead1
phead1 = phead1.next
else:
rs.next = phead2
phead2 = phead2
rs = rs.next
while phead1:
rs.next = phead1
while phead2:
rs.next = phead2
return head
class solution:
# 返回 randomlistnode
def clone(self, phead):
if not phead:return none
p = phead
# 完成節點複製
while p:
node = randomlistnode(p.label)
tail = p.next
p.next = node
node.next=tail
p = tail
# 複製隨機指標
q = phead
while q:
if q.random:
q.next.random = q.random.next
q = q.next
# 開始連線複製的節點
clonode = phead
phead = phead.next
while clonode.next:
node = clonode.next
clonode.next = node.next
clonode = node
return phead
class solution:
def findkthtotail(self, head, k):
# write code here
if not head:return head
p = head
cnt = 0
while p:
cnt += 1
p = p.next
if k>cnt:return
n = cnt-k+1
dummy = listnode(-1)
dummy.next = head
q = dummy
cnt = 0
while q.next:
cnt += 1
if cnt==n:
return q.next.val
q = q.next
class solution:
def entrynodeofloop(self, phead):
p1 = phead #slow
p2 = phead #fast
meet = phead
flag = 0
while p2 and p2.next:
p1 = p1.next
p2 = p2.next.next
if p1 == p2:
flag = 1
meet = p1
break
if not flag:
return none
start = phead
while meet and start:
if meet == start:
return start
meet = meet.next
start = start.next
return none
《劍指offer 之鍊錶題目
這幾天在看 劍指offer 寫的很好，跟july的各有千秋，更加注重歸納總結，筆試面試臨時抱佛腳，決定要啃下這其中的46道題目。ok，把每道題目自己實現，相同的歸成一類。這次把所有的與list相關的題目列出來。思路 從尾到頭列印list的話，考慮用stack先把各節點儲存起來，遍歷完以後再將stac...
劍指Offer題目1518 反轉鍊錶
題目1518 反轉鍊錶 時間限制 1 秒 記憶體限制 128 兆 特殊判題 否 提交 3300 解決 1207 題目描述 輸入乙個鍊錶，反轉鍊錶後，輸出鍊錶的所有元素。hint 請務必使用鍊錶 輸入 輸入可能包含多個測試樣例，輸入以eof結束。對於每個測試案例，輸入的第一行為乙個整數n 0 n 10...
劍指offer 鍊錶
單向鍊錶的結構定義 typedef int datatype struct listnode 問題1 往鍊錶的末尾新增乙個結點 給定頭結點，往末尾插入乙個結點 void insertnode listnode head,datatype key listnode p head while p nex...