[LeetCode] Algorithms-2. Add Two Numbers

描述

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

思路

参考 链接

代码

使用c++的话要用到指针的赋值，相比java会麻烦一点，其他的都是一样的

/* *java实现 */class Solution {    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        int carry  = 0;        ListNode a = l1;         ListNode b = l2;        ListNode result = new ListNode(0);        ListNode head = result;        while(a != null || b != null || carry != 0) {            int x = 0, y = 0, s;            if(a != null) {                x = a.val;                a = a.next;            }            if(b != null) {                y = b.val;                b = b.next;            }            s = x + y + carry;            carry = s / 10;            head.next = new ListNode(s % 10);            head = head.next;        }        return result.next;    }}

/* *c++ 实现 */class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {    int carry = 0;    ListNode *head = new ListNode(0);    ListNode *result = head;    while(l1 != NULL || l2 != NULL || carry != 0) {        int x = 0, y =0, s;        if(l1 != NULL)         {            x = l1->val;             l1 = l1->next;        }        if(l2 != NULL) {            y = l2->val;            l2 = l2->next;        }        s = x + y + carry;        carry = s / 10;        head->next = new ListNode(s % 10);        head = head->next;    }    return result->next; }};