【Leetcode】002 Add Two Numbers

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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

思路

使用dummyHead避免写重复的代码，非常巧妙

实现

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode addTwoNumbers0(ListNode l1, ListNode l2) {        int val = (l1.val + l2.val) % 10;        int increment = (l1.val + l2.val) / 10;        ListNode head = new ListNode(val);        ListNode currNode = head;        l1 = l1.next;        l2 = l2.next;        for (ListNode n1 = l1, n2 = l2; l1 != null && l2 != null; l1 = l1.next, l2 = l2.next){            int result = l1.val + l2.val + increment;            currNode.next = new ListNode(result % 10);            increment = result / 10;            currNode = currNode.next;        }        for (; l1 != null; l1 = l1.next){            int result = l1.val + increment;            currNode.next = new ListNode(result % 10);            increment = result / 10;            currNode = currNode.next;        }        for (; l2 != null; l2 = l2.next){            int result = l2.val + increment;            currNode.next = new ListNode(result % 10);            increment = result / 10;            currNode = currNode.next;        }        if (increment == 1){            currNode.next = new ListNode(1);        }        return head;    }    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        ListNode dummyHead = new ListNode(0);  // 第二个结点是链表的头结点        int increment = 0;        ListNode currNode = dummyHead;        for (ListNode n1 = l1, n2 = l2; l1 != null || l2 != null;){            int x = l1 != null ? l1.val : 0;            int y = l2 != null ? l2.val : 0;            int result = x + y + increment;            currNode.next = new ListNode(result % 10);            increment = result / 10;            currNode = currNode.next;            if (l1 != null) l1 = l1.next;            if (l2 != null) l2 = l2.next;        }        if (increment == 1){            currNode.next = new ListNode(1);        }        return dummyHead.next;    }}