【leetcode】2. Add Two Numbers（Python & C++）

2. Add Two Numbers

题目链接

2.1 题目描述：

You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)

Output: 7 -> 0 -> 8

2.2 解题思路：

思路一：同时遍历两个链表，并把遍历的两个元素相加，创建新链表元素，记录和。如果其中一个链表遍历结束，则新链表直接连接另一个链表剩余元素。最后再处理新链表每个元素是否大于10，要进位的问题。

思路二：同思路一相同，但不同的是在遍历两个元素相加的时候，就判断是否要进位的问题，并用flag记录。这里要注意的是，当两个链表都遍历完，且最后一次计算flag为1，需要进位，则直接创建值为1的新节点连接到链表后面。

思路三：初始化sum值为0。不同于前两种思路遍历条件并不是两个链表非空，而是其中之一不空即可。进入到遍历时，将sum/10，获取进位值。然后依次判断链表是否为空，不空则其值依次与sum相加。新链表的节点值为sum%10。最后在遍历结束后，再判断依次sum/10，若为1，则创建值为1的节点连接到新链表后。

思路四：同思路三一样，不过写法上更加简洁。

2.3 C++代码：

1、思路一代码（72ms）：

class Solution91 {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
if (l1 == NULL && l2 == NULL)
return NULL;
if (l1 == NULL)
return l2;
if (l2 == NULL)
return l1;
ListNode *p1 = l1;
ListNode *p2 = l2;
ListNode *q = new ListNode(-1);
ListNode *p = q;
while (p1!=NULL && p2!=NULL)
{
ListNode *temp = new ListNode(p1->val + p2->val);
p->next = temp;
p = p->next;
p1 = p1->next;
p2 = p2->next;
}
if (p1!=NULL)
p->next = p1;
if (p2 != NULL)
p->next = p2;
int flag = 0;
ListNode *t = q->next;
while (t!=NULL)
{
t->val = t->val + flag;
if (t->val>= 10)
{
t->val = t->val % 10;
flag = 1;
}
else
flag = 0;
if (flag == 1 && t->next==NULL)
{
ListNode *m = new ListNode(1);
t->next = m;
break;
}
t = t->next;
}
return q->next;
}
};

2、思路二代码（46ms）：

class Solution91_1 {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
if (l1 == NULL && l2 == NULL)
return NULL;
if (l1 == NULL)
return l2;
if (l2 == NULL)
return l1;
ListNode *p1 = l1;
ListNode *p2 = l2;
ListNode *q = new ListNode(-1);
ListNode *p = q;
int flag = 0;
while (p1 != NULL && p2 != NULL)
{
ListNode *temp = new ListNode(p1->val + p2->val + flag);
if (temp->val >= 10)
{
temp->val = temp->val % 10;
flag = 1;
}
else
flag = 0;
p->next = temp;
p = p->next;
if (flag == 1 && p1->next == NULL && p2->next==NULL)
{
ListNode *x = new ListNode(1);
p->next = x;
}
p1 = p1->next;
p2 = p2->next;
}
while (p1 != NULL)
{
p1->val = p1->val + flag;
if (p1->val >= 10)
{
p1->val = p1->val % 10;
flag = 1;
}
else
flag = 0;
p->next = p1;
p = p->next;
if (flag == 1 && p1->next==NULL)
{
ListNode *x = new ListNode(1);
p->next = x;
break;
}
p1 = p1->next;
}
while (p2 != NULL)
{
p2->val = p2->val + flag;
if (p2->val >= 10)
{
p2->val = p2->val % 10;
flag = 1;
}
else
flag = 0;
p->next = p2;
p = p->next;
if (flag == 1 && p2->next == NULL)
{
ListNode *x = new ListNode(1);
p->next = x;
break;
}
p2 = p2->next;
}
return q->next;
}
};

3、思路三代码（32ms）：

class Solution91_2 {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
if (l1 == NULL && l2 == NULL)
return NULL;
if (l1 == NULL)
return l2;
if (l2 == NULL)
return l1;
ListNode *p1 = l1;
ListNode *p2 = l2;
ListNode *q = new ListNode(-1);
ListNode *p = q;
int sum = 0;
while (p1 != NULL || p2 != NULL)
{
sum = sum / 10;
if (p1 != NULL)
{
sum = sum + p1->val;
p1 = p1->next;
}
if (p2 != NULL)
{
sum = sum + p2->val;
p2 = p2->next;
}
p->next = new ListNode(sum % 10);
p = p->next;
}
if (sum / 10 == 1)
p->next = new ListNode(1);
return q->next;
}
};

4、思路四代码（29ms）：

class Solution91_3 {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
if (l1 == NULL && l2 == NULL)
return NULL;
if (l1 == NULL)
return l2;
if (l2 == NULL)
return l1;
ListNode *p1 = l1;
ListNode *p2 = l2;
ListNode *q = new ListNode(-1);
ListNode *p = q;
int flag = 0;
while (p1 || p2 || flag)
{
int sum = (p1 ? p1->val : 0) + (p2 ? p2->val : 0) + flag;
p->next = new ListNode(sum % 10);
p = p->next;
flag = sum / 10;
p1 = p1 ? p1->next : p1;
p2 = p2 ? p2->next : p2;
}
return q->next;
}
};

2.4 Python代码：

1、思路二代码（115ms）：

class Solution(object):
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
if l1==None and l2==None:
return None
if l1==None:
return l2
if l2==None:
return l1
q=ListNode(-1)
p=q
flag=0
p1=l1
p2=l2
while p1!=None and p2!=None:
temp=ListNode(p1.val+p2.val+flag)
if temp.val>=10:
temp.val=temp.val%10
flag=1
else:
flag=0
p.next=temp
p=p.next
if flag==1 and p1.next==None and p2.next==None:
t=ListNode(1)
p.next=t
p1=p1.next
p2=p2.next
while p1!=None:
p1.val=p1.val+flag
if p1.val>=10:
p1.val=p1.val%10
flag=1
else:
flag=0
p.next=p1
p=p.next
if flag==1 and p1.next==None:
t=ListNode(1)
p.next=t
break
p1=p1.next
while p2!=None:
p2.val=p2.val+flag
if p2.val>=10:
p2.val=p2.val%10
flag=1
else:
flag=0
p.next=p2
p=p.next
if flag==1 and p2.next==None:
t=ListNode(1)
p.next=t
break
p2=p2.next
return q.next

2、思路三代码（132ms）：

class Solution1(object):
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
if l1==None and l2==None:
return None
if l1==None:
return l2
if l2==None:
return l1
q=ListNode(-1)
p=q
flag=0
p1=l1
p2=l2
while p1 or p2 or flag:
sum=flag
if p1:
sum=sum+p1.val
p1=p1.next
if p2:
sum=sum+p2.val
p2=p2.next
p.next=ListNode(sum%10)
p=p.next
flag=sum/10
return q.next