2020-11-22 2. Add Two Numbers

Medium

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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 contains a single digit. Add the two numbers and return the sum as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example 1:

Input:l1 = [2,4,3], l2 = [5,6,4]Output:[7,0,8]Explanation:342 + 465 = 807.

Example 2:

Input:l1 = [0], l2 = [0]Output:[0]

Example 3:

Input:l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]Output:[8,9,9,9,0,0,0,1]

Constraints:

The number of nodes in each linked list is in the range [1, 100].

0 <= Node.val <= 9

It is guaranteed that the list represents a number that does not have leading zeros.

/**

* Definition for singly-linked list.

* public class ListNode {

*    int val;

*    ListNode next;

*    ListNode() {}

*    ListNode(int val) { this.val = val; }

*    ListNode(int val, ListNode next) { this.val = val; this.next = next; }

* }

*/

class Solution {

    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {           

        // 进位

        int carry = 0;

        // 结果，当前位

        ListNode res, cur;

        res = cur = new ListNode(0);

        // 两个数的当前位

        ListNode n1 = l1, n2 = l2;

        while (n1 != null || n2 != null || carry != 0) {

            // 当前位数相加，并加上进位

            cur.val = (n1 != null ? n1.val : 0) + (n2 != null ? n2.val : 0) + carry;

            // 清除进位

            carry = 0;

            // 如果当前产生了进位，则位数取个位，然后设置进位

            if (cur.val >= 10) { cur.val -= 10; carry = 1; }

            // 处理下一位

            if (n1 != null) n1 = n1.next;

            if (n2 != null) n2 = n2.next;

            if (n1 != null || n2 != null || carry != 0) cur = cur.next = new ListNode(0);

        }

        return res;

    }

}