leetcode——2

1. 题目

Add Two Numbers

You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)   
Output: 7 -> 0 -> 8        即342＋465=807

给你两个链表代表两个非负数。数字以相反的顺序存储，每个节点包含一个单一的数字。加上这两个数并返回一个链表。

2.c++解题

 //LeetCode_Add Two Numbers//Written by zhou//2013.11.1

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.                if (l1 == NULL) return l2;        if (l2 == NULL) return l1;

        ListNode *resList = NULL, *pNode = NULL, *pNext = NULL;    // resList头节点， pNode 每轮的末节点，  pNext临时节点        ListNode *p = l1, *q = l2;        int up = 0;        while(p != NULL && q != NULL)        {            pNext = new ListNode(p->val + q->val + up);            up = pNext->val / 10;    //计算进位            pNext->val = pNext->val % 10;   //计算该位的数字                        if (resList == NULL)  //头结点为空            {                resList = pNode = pNext;            }            else //头结点不为空            {                pNode->next = pNext;                pNode = pNext;            }            p = p->next;            q = q->next;        }

        //处理链表l1剩余的高位        while (p != NULL)        {            pNext = new ListNode(p->val + up);            up = pNext->val / 10;                pNext->val = pNext->val % 10;            pNode->next = pNext;            pNode = pNext;            p = p->next;        }

        //处理链表l2剩余的高位        while (q != NULL)        {            pNext = new ListNode(q->val + up);            up = pNext->val / 10;                pNext->val = pNext->val % 10;            pNode->next = pNext;            pNode = pNext;            q = q->next;        }

        //如果有最高处的进位，需要增加结点存储        if (up > 0)        {            pNext = new ListNode(up);            pNode->next = pNext;        }

        return resList;    }

};

3. python解题

3.1

# Definition for singly-linked list.# class ListNode:#     def __init__(self, x):#         self.val = x#         self.next = None           class Solution:    # @return a ListNode    def addTwoNumbers(self, l1, l2)：        dummy, flag = ListNode(0), 0        head = dummy        while flag or l1 or l2:       ／／flag 进位，  node临时节点， dummy最后节点            node = ListNode(flag)            if l1:                node.val += l1.val                l1 = l1.next            if l2:                node.val += l2.val                l2 = l2.next            flag = node.val / 10            node.val %= 10            head.next = node              head = head.next  # head.next, head = node, node        return dummy.next

3.2

# Definition for singly-linked list.# class ListNode:#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution:    # @return a ListNode    def addTwoNumbers(self, l1, l2):　　　　 if not l1: return l2        if not l2: return l1        dummy = ListNode(0)        p = dummy        flag = 0        while l1 and l2:            tmp = l1.val + l2.val + flag            p.next = ListNode( tmp % 10 )            flag = tmp / 10            l1, l2, p = l1.next, l2.next, p.next        if l1:            while l1:                tmp = l1.val + flag                p.next = ListNode( tmp % 10 )                flag = tmp / 10                l1, p = l1.next, p.next        if l2:            while l2:                tmp = l2.val + flag                p.next = ListNode( tmp % 10 )                flag = tmp / 10                l2, p = l2.next, p.next        if flag == 1: p.next = ListNode(flag)        return dummy.next