LinkedList - 2. Add Two Numbers

2. Add Two Numbers

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.

思路：

题目意思是给两个链表，把这两个链表，一个链表代表一个数，不过顺序是从链表尾数到链表头，然后需要把两个数加起来，形成一个新的链表，这个新的链表也是从倒序的。做法很简单，就是从头开始遍历两个链表，依次相加然后模10操作得到当前节点的Value，如果超过10，就向后传递一个1。最后需要注意，如果两个链表都结束，但是最后一位被进位之后还超过10，需要额外添加一个1节点，比如最后一个value是9，前面依次进位，进到9之后变成10这种情况。

代码：

java：

/**

 * 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(0);
        ListNode curr = dummy;
        int sum = 0;
        while (l1 != null || l2 != null) {
            if (l1 != null) {
                sum += l1.val;
                l1 = l1.next;
            }
            
            if (l2 != null) {
                sum += l2.val;
                l2 = l2.next;
            }
            
            curr.next = new ListNode(sum % 10);
            sum /= 10;
            curr = curr.next;
        }
        
        if (sum == 1) {
            curr.next = new ListNode(1);
        }
        
        return dummy.next;
    }
}

go：

/**

 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
    var dummy = &ListNode{Val:-1};
    var p = dummy

    var sum int
    for l1 != nil || l2 != nil {
        if l1 != nil {
            sum += l1.Val
            l1 = l1.Next
        }
        if l2 != nil {
            sum += l2.Val
            l2 = l2.Next
        } 

        p.Next = &ListNode{Val:sum % 10}
        p = p.Next
        sum /= 10
    }

    if sum == 1 {
        p.Next = &ListNode{Val:sum}
    }

    return dummy.Next
}