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 = Noneclass 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 listclass 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_listif __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

这个函数的功能是整除，返回值为（进位，余数），这也比我单独设置carry好

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