#!/usr/bin/env python
# -*- coding:utf-8 -*-
class
listnode
(object):
def__init__
(self,val=
none):
self.val = val
self.
next
=none
class
listnode_handle
:# 定義單鏈表的各種操作
def__init__
(self)
: self.cur_node =
none
# self.nodelist=none # 用來儲存結果的鍊錶,方法3?????
defadd
(self, data)
:# 頭插法(每次新增新節點都是在首節點之後新增)
node = listnode(
) node.val = data
node.
next
= self.cur_node
self.cur_node = node # cur_node總是指向新節點
return node
defprint_listnode
(self, node)
:while node:
# 當條件變成node.val時可能報錯,最後乙個元素可能是空節點,它沒有資料域和指標域
# print ('\nnode: ', node, ' value: ', node.val, ' next: ', node.next)
print
(node.val)
node = node.
next
defget_list
(self, nodelist)
:# 能否實現利用for 迴圈快速遍歷
list=[
]while nodelist:
list
nodelist = nodelist.
next
return
list
def__next__
(self,node)
:# 奇思妙想:就是要遍歷當前node節點後所有的元素
pass
class
solution
(object):
defaddtwonumbers
(self, l1, l2)
: carry =
0 result_handle = listnode_handle(
) result = listnode(
)while l1 !=
none
or l2 !=
none:if
not carry:
# 無進製
if l1 !=
none
and l2 !=
none
: l3_val = l1.val + l2.val
elif l1 ==
none
and l2 !=
none
: l3_val = l2.val
elif l1 !=
none
and l2 ==
none
: l3_val = l1.val
else
:pass
else
:# 有進製
if l1 !=
none
and l2 !=
none
: l3_val = l1.val + l2.val +
1elif l1 ==
none
and l2 !=
none
: l3_val = l2.val +
1elif l1 !=
none
and l2 ==
none
: l3_val = l1.val +
1else
:pass
carry =
0# 處理完進製後,將進製標誌置空
if l3_val <10:
result = result_handle.add(l3_val)
else
: result = result_handle.add(l3_val %10)
carry = l3_val //
10try
: l1 = l1.
next
except
:pass
try:
l2 = l2.
next
# l2 為none,執行except語句
except
:pass
if carry:
result = result_handle.add(carry)
# result_handle.print_listnode(result)
return result
class
solution1
:# @return a listnode
defaddtwonumbers
(self, l1, l2)
: carry =
0 root = n = listnode(
'prior'
)# root 始終指向第乙個頭結點,n節點則是指向新加入的節點
while l1 or l2 or carry:
# 防止l1 or l2 為空鍊錶
v1 = v2 =
0if l1:
v1 = l1.val
l1 = l1.
next
if l2:
v2 = l2.val
l2 = l2.
next
carry, val =
divmod
(v1 + v2 + carry,
10)
n.next
= listnode(val)
# 尾插法新增新節點
n = n.
next
return root.
next
# 結果時逆序的
if __name__ ==
"__main__"
: listnode_1 = listnode_handle(
) l1 = listnode(
) l1_list =[9
,9]# 利用不同的listnode_handle類進行處理,
# 因為使用頭插法插入節點,正好每個節點都是倒序鏈結起來,方便了遍歷鍊錶結算每位數的相加
for i in l1_list:
l1 = listnode_1.add(i)
listnode_2 = listnode_handle(
) l2 = listnode(
) l2_list =[1
]for i in l2_list:
l2 = listnode_2.add(i)
# result = solution().addtwonumbers(l1, l2)
# result1_node1 = listnode_handle()
# des=result1_node1.get_list(result)
# 方法2 構建
result1 = solution1(
).addtwonumbers(l1, l2)
result1_node = listnode_handle(
) des=result1_node.get_list(result1)
print
(des[::
-1])
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給定兩個非空鍊錶來表示兩個非負整數。位數按照逆序方式儲存，它們的每個節點只儲存單個數字。將兩數相加返回乙個新的鍊錶。你可以假設除了數字 0 之外，這兩個數字都不會以零開頭。示例 輸入 2 4 3 5 6 4 輸出 7 0 8原因 342 465 807 definition for singly l...