solution:
需要考虑进位的问题。 然后就是linked list细节了: 生成新节点等等。
代码:
/**
* 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(-1);
ListNode* p = &dummy;
int carrier = 0;
while(l1||l2){
if(l1) {
carrier += l1->val;
l1 = l1->next;
}
if(l2) {
carrier += l2->val;
l2 = l2->next;
}
p->next = new ListNode(carrier%10);
p = p->next;
carrier /= 10;
}
if(carrier){
p->next = new ListNode(carrier);
p = p->next;
}
return dummy.next;
};
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