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 = None
class 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