實現鏈表形式的兩數相加 Add Two Numbers

問題：

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

解決：

① 按照正常的數字相加，從左往右加便可。須要注意的是使用ListNode保存結果，因此要處理好。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution { //56ms
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode res = new ListNode(-1);
        ListNode p1 = l1,p2 = l2,cur = res;
        int carry = 0;
        while(p1 != null || p2 != null){
            int sum = 0;
            if(p1 != null){
                sum += p1.val;
                p1 = p1.next;
            }
            if(p2 != null){
                sum += p2.val;
                p2 = p2.next;
            }
            sum += carry;
            cur.next = new ListNode(sum % 10);
            cur = cur.next;
            carry = sum / 10;
        }
        if(carry == 1) cur.next = new ListNode(1);
        return res.next;
    }
}

② 進化版

public class Solution { //52ms
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode res = new ListNode(0);
        ListNode p = l1, q = l2, cur = res;
        int carry = 0;
        while (p != null || q != null) {
            int x = (p != null) ? p.val : 0;
            int y = (q != null) ? q.val : 0;
            int sum = carry + x + y;
            carry = sum / 10;
            cur.next = new ListNode(sum % 10);
            cur = cur.next;
            if (p != null) p = p.next;
            if (q != null) q = q.next;
        }
        if (carry > 0) {
            cur.next = new ListNode(carry);
        }
        return res.next;
    }
}