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)
代码
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* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* (struct ListNode* l1, struct ListNode* l2)
struct ListNode* c1 = l1;
struct ListNode* c2 = l2;
struct ListNode* sentinel = (struct ListNode*)malloc (sizeof (struct ListNode));
struct ListNode* dummy = sentinel;
int sum = 0 ;
while ( c1 || c2) {
sum /= 10 ;
if ( c1 ){
sum += c1 -> val;
c1 = c1 -> next;
}
if ( c2 ){
sum += c2 -> val;
c2 = c2 -> next;
}
dummy -> next = (struct ListNode*)malloc (sizeof (struct ListNode));
dummy -> next -> val = sum % 10 ;
dummy = dummy -> next;
}
if ( sum / 10 == 1 ) {
dummy -> next = (struct ListNode*)malloc (sizeof (struct ListNode));
dummy -> next -> val = 1 ;
}
return sentinel -> next;
}
C++版本：
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* 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 preheader (-1 ) , *curr =&preheader;
int carry=0 ;
while (l1 || l2 || carry) {
curr->next = new ListNode(((l1?l1->val:0 )+(l2?l2->val:0 )+carry)%10 );
curr = curr->next;
carry = ((l1?l1->val:0 )+(l2?l2->val:0 )+carry)/10 ;
l1?l1=l1->next:0 ;
l2?l2=l2->next:0 ;
}
return preheader.next;
}
};
1558 / 1558 test cases passed.Status: Accepted40 ms
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