2. Add Two Numbers 【Medium】

avatar 2020年08月29日14:30:57 6 2303 views
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Question

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.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

My Answer

/**
 * 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) {
        ListNode c1 = l1;
        ListNode c2 = l2;
        ListNode head = new ListNode(0); // 虚拟头结点
        ListNode curr = head;
        int carry = 0;
        while(c1 != null || c2 != null) {
            int x = (c1 != null) ? c1.val : 0;
            int y = (c2 != null) ? c2.val : 0;
            int sum = carry + x + y;
            carry = sum / 10;
            
            curr.next = new ListNode(sum % 10);
            curr = curr.next;
            
            if(c1 != null) {
                c1 = c1.next;
            }
            if(c2 != null) {
                c2 = c2.next;
            }
        }
        
        // 解决最后一个结点之间的相加进位的情况
        if(carry > 0) {
            curr.next = new ListNode(carry);
        }
        return head.next;
    }
}
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