LeetCode 2：Add Two Numbers

最新推荐文章于 2020-07-14 12:02:11 发布

weixin_30701521

最新推荐文章于 2020-07-14 12:02:11 发布

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/whuyt/p/4883846.html

本文介绍了如何通过引入虚拟头结点简化链表加法操作的实现过程，以解决节点是否为空的问题，使得代码逻辑更加清晰简洁。

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public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
       ListNode cur1 = l1;
       ListNode cur2 = l2;
       ListNode head = null;
       ListNode cur = null;
       int sum = 0;
       while (cur1 != null || cur2 != null) {
           if (cur1 != null) {
               sum += cur1.val;
               cur1 = cur1.next;
           }
           if (cur2 != null) {
               sum += cur2.val;
               cur2 = cur2.next;
           }
           if (head == null) {
               head = new ListNode(sum % 10);
               cur = head;
           } else {
               cur.next = new ListNode(sum % 10);
               cur = cur.next;
           }
           sum /= 10;
       }
       if (sum == 1)
           cur.next = new ListNode(1);
       return head;
    }
}

中间那部分处理head是否为null的情况比较麻烦，可以采用一个 dummy node 简化

public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
       ListNode cur1 = l1;
       ListNode cur2 = l2;
       ListNode head = new ListNode(0);
       ListNode cur = head;
       int sum = 0;
       while (cur1 != null || cur2 != null) {
           if (cur1 != null) {
               sum += cur1.val;
               cur1 = cur1.next;
           }
           if (cur2 != null) {
               sum += cur2.val;
               cur2 = cur2.next;
           }
           cur.next = new ListNode(sum % 10);
           cur = cur.next;
           sum /= 10;
       }
       if (sum == 1)
           cur.next = new ListNode(1);
       return head.next;
    }
}