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
这道题是计算用链表表示的两个数字之和,题目难度为Medium。
题目比较简单,注意进位的处理即可,就不详细说明了,代码中没有释放内存,具体项目时相信大家都会注意。具体代码:
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
int carry = 0;
ListNode* head = NULL;
ListNode* prev = NULL;
while(l1 || l2 || carry) {
int num1 = l1 ? l1->val : 0;
int num2 = l2 ? l2->val : 0;
int sum = num1 + num2 + carry;
ListNode* node = new ListNode(sum%10);
if(!head) head = node;
if(prev) prev->next = node;
prev = node;
carry = sum / 10;
if(l1) l1 = l1->next;
if(l2) l2 = l2->next;
}
return head;
}
};