题目:
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
有两个链表作为输入,它们表示逆序的两个非负数。如下面的两个链表表示的是342和465这两个数。你需要计算它们的和并且用同样的方式逆序输出。
除了0之外,两个数首位均不为0.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
思路:
指针遍历两个链表,用一个变量表示进位,指向两个链表的指针有一个为空或进位为空时终止循环。
代码:
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode* head=new ListNode(0);
ListNode* ptr=head;
int carry = 0;
while(true){
if(l1!=NULL){
carry+=l1->val;
l1=l1->next;
}
if(l2!=NULL){
carry+=l2->val;
l2=l2->next;
}
ptr->val=carry%10;
carry/=10;
if(l1!=NULL||l2!=NULL||carry!=0){
ptr=(ptr->next=new ListNode(0));
}
else
break;
}
return head;
}
};
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode* head=new ListNode(0);
ListNode* ptr=head;
int carry = 0;
while(true){
if(l1!=NULL){
carry+=l1->val;
l1=l1->next;
}
if(l2!=NULL){
carry+=l2->val;
l2=l2->next;
}
ptr->val=carry%10;
carry/=10;
if(l1!=NULL||l2!=NULL||carry!=0){
ptr=(ptr->next=new ListNode(0));
}
else
break;
}
return head;
}
};