leetcode-Add Two Numbers

最新推荐文章于 2024-08-12 10:14:14 发布

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本文介绍如何使用链表实现两个非负整数相加，并返回结果作为新的链表。通过遍历两个链表并考虑进位，实现两数之和的计算。

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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

思路较简单，主要是注意进位等情况

/**
* 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) {
if(l1==NULL)
{
return l2;
}
else if(l2==NULL)
{
return l1;
}
else
{
int carry=0;
int temp=0;
ListNode *p1=l1;
ListNode *p2=l2;
ListNode *pre;
while(( p1!=NULL ) && ( p2!=NULL ))
{
temp=p1->val+p2->val+carry;
p1->val=temp%10;
if (temp>9)
{
carry=temp/10;
temp=0;
if((p1->next==NULL)&&(p2->next==NULL))
{
ListNode *newNode=(ListNode *)malloc(sizeof(ListNode));
newNode->val=carry;
newNode->next=NULL;
p1->next=newNode;
return l1;
}
}
else
{
carry=0;
temp=0;
if((p1->next==NULL)&&(p2->next==NULL))
{

return l1;
}
}
pre=p1;
p1=p1->next;
p2=p2->next;
}
if(p2==NULL)
{
while(p1!=NULL)
{
temp=p1->val+carry;
p1->val=temp%10;
if (temp>9)
{
carry=temp/10;
temp=0;
if(p1->next==NULL)
{
ListNode *newNode=(ListNode *)malloc(sizeof(ListNode));
newNode->val=carry;
newNode->next=NULL;
p1->next=newNode;
return l1;
}
}
else
{
carry=0;
temp=0;
if(p1->next==NULL)
{
return l1;
}
}
p1=p1->next;
}
}
else if(p1==NULL)
{
while(p2!=NULL)
{
temp=p2->val+carry;
ListNode *newNode=(ListNode *)malloc(sizeof(ListNode));
newNode->val=temp%10;
newNode->next=NULL;
pre->next=newNode;
pre=newNode;
if (temp>9)
{
carry=temp/10;
temp=0;
if(p2->next==NULL)
{
ListNode *newNode=(ListNode *)malloc(sizeof(ListNode));
newNode->val=carry;
newNode->next=NULL;
pre->next=newNode;
return l1;
}
}
else
{
carry=0;
temp=0;
if(p2->next==NULL)
{
return l1;
}
}

p2=p2->next;
}
}
}
}
};