leetcode: 2 Add_Two_Numbers
第二题,没什么复杂的直接上代码好了,简单说就是首先两个链表相加,然后反向存储,最后转化成list输出
#!/usr/bin/env python
# -*- coding:utf-8 -*-
"""
Copyright (c) 2017 Xu Zhihao (Howe). All rights reserved.
This program is free software; you can redistribute it and/or modify
"""
carray = False
# Definition for singly-linked list.
class ListNode(object):
def __init__(self):
self.val = None
self.next = None
class ListNode_handle:
def __init__(self):
self.cur_node = None
def add(self, data):
#add a new node pointed to previous node
node = ListNode()
node.val = data
node.next = self.cur_node
self.cur_node = node
return node
def print_ListNode(self, node):
while node:
print '\nnode: ', node, ' value: ', node.val, ' next: ', node.next
node = node.next
def _reverse(self, nodelist):
list = []
while nodelist:
list.append(nodelist.val)
nodelist = nodelist.next
result = ListNode()
result_handle = ListNode_handle()
for i in list:
result = result_handle.add(i)
return result
def get_list(self, nodelist):
list = []
while nodelist:
list.append(nodelist.val)
nodelist = nodelist.next
return list
class Solution(object):
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
global carry
carry = False
result_handle = ListNode_handle()
result = ListNode()
while l1 != None or l2 != None:
if not carry:
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:
if l1 != None and l2 != None:
l3_val = l1.val + l2.val + 1
elif l1 == None and l2 != None:
l3_val = l2.val + 1
elif l1 != None and l2 == None:
l3_val = l1.val + 1
else :
pass
carry = False
if l3_val < 10:
result = result_handle.add(l3_val)
else:
result = result_handle.add(l3_val%10)
carry = True
try:
l1 = l1.next
except:
pass
try:
l2 = l2.next
except:
pass
result = result_handle._reverse(result)
result_list = []
result_list = result_handle.get_list(result)
if carry:
result_list.append(1)
result = result_handle.add(1)
result = result_handle._reverse(result)
result_handle.print_ListNode(result)
return result_list
if __name__ == "__main__":
# creat 2 linked lists
ListNode_1 = ListNode_handle()
l1 = ListNode()
l1_list = [1,8]
for i in l1_list:
l1 = ListNode_1.add(i)
ListNode_2 = ListNode_handle()
l2 = ListNode()
l2_list = [9]
for i in l2_list:
l2 = ListNode_2.add(i)
#reverse 2 linked lists
l1 = ListNode_1._reverse(l1)
l2 = ListNode_2._reverse(l2)
#get result
result = Solution().addTwoNumbers(l1, l2)
print result首先设置进位:
carray = False然后设置链表格式:
# Definition for singly-linked list.
class ListNode(object):
def __init__(self):
self.val = None
self.next = None然后再设置添加链表功能:
def add(self, data):
#add a new node pointed to previous node
node = ListNode()
node.val = data
node.next = self.cur_node
self.cur_node = node
return node设置反向:
def _reverse(self, nodelist):
list = []
while nodelist:
list.append(nodelist.val)
nodelist = nodelist.next
result = ListNode()
result_handle = ListNode_handle()
for i in list:
result = result_handle.add(i)
return result输出list
def get_list(self, nodelist):
list = []
while nodelist:
list.append(nodelist.val)
nodelist = nodelist.next
return list两个链表相加的结果函数:
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
global carry
carry = False
result_handle = ListNode_handle()
result = ListNode()
while l1 != None or l2 != None:
if not carry:
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:
if l1 != None and l2 != None:
l3_val = l1.val + l2.val + 1
elif l1 == None and l2 != None:
l3_val = l2.val + 1
elif l1 != None and l2 == None:
l3_val = l1.val + 1
else :
pass
carry = False
if l3_val < 10:
result = result_handle.add(l3_val)
else:
result = result_handle.add(l3_val%10)
carry = True
try:
l1 = l1.next
except:
pass
try:
l2 = l2.next
except:
pass
result = result_handle._reverse(result)
result_list = []
result_list = result_handle.get_list(result)
if carry:
result_list.append(1)
result = result_handle.add(1)
result = result_handle._reverse(result)
result_handle.print_ListNode(result)
return result_list官网上的解：
class Solution(object):
def addTwoNumbers(self, l1, l2):
carry = 0
root = n = ListNode(0)
while l1 or l2 or carry:
v1 = v2 = 0
if 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)
n = n.next
return root.next分析：
这里他默认了l1，l2是链表了，所以我之前建立链表的操作比较多余。
然后算法上，他用了
divmod整体思路是：
先定义了root 和 n同时候一个指向None的值为0的ListNode，root不变
然后遍历l1，l2,
如过l1有值则取出l1中的值v1，然后将l1指向下一个
如过l2有值则取出l2中的值v2，然后将l2指向下一个
接着计算（值v1+值v2+进位）的求和之后对10求余数，得到新的进位和余数
再接着，将n指向新的余数，root保持不变
依次循环
返回root的所指向的下一位
eg：(2 -> 4 -> 3) + (5 -> 6 -> 4)
他的结果是：7 -> 0 -> 8
但是root的值为0 -> 7 -> 0 -> 8
GitHub: https://github.com/DinnerHowe/LeetCode.git