2. Add Two Numbers

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本文介绍了一种使用链表实现两数相加的方法。通过迭代处理输入的链表节点，并正确处理进位，最终返回结果链表。讨论了不同情况下的处理方式及优化方案。

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题目

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.

我的思路

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode((l1.val + l2.val) % 10);
        int carry = (l1.val + l2.val) / 10;
        ListNode current = dummy;
        l1 = l1.next;
        l2 = l2.next;
        while(l1 != null && l2 != null){
            int val = (l1.val + l2.val + carry) % 10; 
            ListNode newNode = new ListNode(val);
            current.next = newNode;
            carry = (l1.val + l2.val + carry) / 10;
            l1 = l1.next;
            l2 = l2.next;
            current = current.next;
        }
        while(l1 != null){
            int val = (l1.val + carry) % 10;
            ListNode newNode = new ListNode(val);
            current.next = newNode;
            carry = (l1.val + carry) / 10;
            l1 = l1.next;
            current = current.next;
        }
        while(l2 != null){
            int val = (l2.val + carry) % 10;
            ListNode newNode = new ListNode(val);
            current.next = newNode;
            carry = (l2.val + carry) / 10;
            l2 = l2.next;
            current = current.next;
        }
        if(carry == 1){
            ListNode newNode = new ListNode(1);
            current.next = newNode;
            current = current.next;
        }
        return dummy;
    }
}

2019.11.7 update:
之前的写法可太傻逼了

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode carry = new ListNode(0);
        ListNode dummy = new ListNode(0);
        ListNode head = dummy;
        while(l1 != null || l2 != null || carry.val != 0) {
            int add1 = l1 == null ? 0 : l1.val;
            int add2 = l2 == null ? 0 : l2.val;
            dummy.next = new ListNode((carry.val + add1 + add2) % 10);
            dummy = dummy.next;
            carry.val = (carry.val + add1 + add2) / 10;
            l1 = l1 == null ? null : l1.next;
            l2 = l2 == null ? null : l2.next;
        }
        return head.next;
    }
}

解答

leetcode solution

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode dummyHead = new ListNode(0);
    ListNode p = l1, q = l2, curr = dummyHead;
    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;
        curr.next = new ListNode(sum % 10);
        curr = curr.next;
        if (p != null) p = p.next;
        if (q != null) q = q.next;
    }
    if (carry > 0) {
        curr.next = new ListNode(carry);
    }
    return dummyHead.next;
}