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.
Example:
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.
My Solution:
/**
* 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) {
ListNode* result=(ListNode*)malloc(sizeof(ListNode));
int carry=0;
ListNode* head=(ListNode*)malloc(sizeof(ListNode));
head=result;
while (l1!= NULL&&l2!= NULL){
ListNode* temp=(ListNode*)malloc(sizeof(ListNode));
temp->val=l1->val+l2->val+carry;
if (temp->val>=10) {
carry=1;
temp->val=temp->val%10;
} else carry=0;
result->next=temp;
result = result->next;
l1=l1->next;
l2=l2->next;
}
while (l1!= nullptr) {
if (carry>0){
ListNode* temp=(ListNode*)malloc(sizeof(ListNode));
temp->val=l1->val+carry;
carry=0;
if (temp->val >= 10 ){
carry=1;
temp->val%=10;
}
result->next=temp;
}
else result->next=l1;
result = result->next;
l1=l1->next;
}
while (l2!= nullptr) {
if (carry>0){
ListNode* temp=(ListNode*)malloc(sizeof(ListNode));
temp->val=l2->val+carry;
carry=0;
if (temp->val >= 10 ){
carry=1;
temp->val%=10;
}
result->next=temp;
}
else result->next=l2;
result = result->next;
l2=l2->next;
}
//最高位进位处理
if (carry>0) {
ListNode* temp=(ListNode*)malloc(sizeof(ListNode));
temp->val=1;
result->next=temp;
result = result->next;
}
result->next= nullptr;
return head->next;
}
};
在LeetCode的评论区网友的指导下,简化了一下代码:
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode* result=(ListNode*)malloc(sizeof(ListNode));
int carry=0;
ListNode* head=(ListNode*)malloc(sizeof(ListNode));
head=result;
while (l1!= NULL||l2!= NULL){
ListNode* temp=(ListNode*)malloc(sizeof(ListNode));
if (l1==NULL)
{
temp->val=l2->val+carry;
}
else if (l2==NULL)
{
temp->val=l1->val+carry;
}else temp->val=l1->val+l2->val+carry;
if (temp->val>=10) {
carry=1;
temp->val=temp->val%10;
} else carry=0;
result->next=temp;
result = result->next;
if (l1!=NULL) l1=l1->next;
if (l2!=NULL) l2=l2->next;
}
if (carry>0) {
ListNode* temp=(ListNode*)malloc(sizeof(ListNode));
temp->val=1;
result->next=temp;
result = result->next;
}
result->next= nullptr;
return head->next;
}
};
总结:非常简单的链表操作;注意处理一下最高位进位的情况就好(i.e. 1+99=100)
在LeetCode的评论区看到一种比较优雅的写法:
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode * l0 = nullptr, * l3 = nullptr;
int a = 0;
while( l1 or l2 or a ) {
if(l1){
a+=l1->val;
l1 = l1->next;
}
if(l2){
a+=l2->val;
l2 = l2->next;
}
if(l3) {
l3->next = new ListNode(a%10);
l3 = l3->next;
}
else {
l3 = l0 = new ListNode(a%10);
}
a /= 10;
}
return l0;
}
};
这个变量a相当巧妙,先保存相加的和,再用%去余数,然后保存进位值,自愧弗如。