Sep 30, 2017

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


ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {

ListNode ret(0);                        //头结点让单链表操作变简单
ListNode *p = &ret;
while(l1 != nullptr && l2 != nullptr){
ans = (l1->val + l2->val) + add;    //记得加上进位
p->next = new ListNode(ans%10);
p = p->next;
l1 = l1->next;
l2 = l2->next;
}
while(l1){
p->next = new ListNode(ans%10);
p = p->next;
l1 = l1->next;
}
while(l2){
p->next = new ListNode(ans%10);
p = p->next;
l2 = l2->next;
}
return ret.next;
}



dicuss中还有一个更精炼的写法：

ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode ret(0);
ListNode *p = &ret;
sum = (l1?l1->val:0) + (l2?l2->val:0) + add;
p->next = new ListNode(sum%10);
p = p->next;
l1 = (l1?l1->next:nullptr);
l2 = (l2?l2->next:nullptr);
}
return ret.next;
}

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