本题是要求把两个链表中的数字一一相加形成一个新的链表,关键在于两个链表可能不一样长,这时可以将短的那个链表剩余的数值全看成0。具体代码如下:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode *head = NULL, *prev = NULL;
int carry = 0;
while (l1 || l2) {
int v1 = l1? l1->val: 0;
int v2 = l2? l2->val: 0;
int tmp = v1 + v2 + carry;
carry = tmp / 10;
int val = tmp % 10;
ListNode* cur = new ListNode(val);
if (!head) head = cur;
if (prev) prev->next = cur;
prev = cur;
l1 = l1? l1->next: NULL;
l2 = l2? l2->next: NULL;
}
if (carry > 0) {
ListNode* l = new ListNode(carry);
prev->next = l;
}
return head;
}