題目:
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
注意:
鏈表l1和l2長度可能不同,要注意處理某個鏈表剩餘的高位;
鏈表上的數相加,要判斷進位是否爲0。
代碼:
/**
* 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 dummy(0);//頭節點
ListNode *p=&dummy;//指針
int carry=0;//進位
while(l1||l2||carry){//只要l1,l2,進位有一不爲0,就要加
int sum=(l1?l1->val:0)+(l2?l2->val:0)+carry;//從個位加起來(因爲數是倒置存儲的)
carry=sum/10;//計算進位
p->next=new ListNode(sum%10);//output,每個節點存儲
p=p->next;//output的下一個節點
l1=l1?l1->next:l1;// l1和l2的下一個節點;
l2=l2?l2->next:l2;
}
return dummy.next;
}
};