Add Two Numbers 2015年6月8日

You are given two linked lists representing two non-negative numbers. 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.

Input: ( ->  -> ) + ( ->  -> )
Output:  ->  -> 

初次看题没懂。

要点是 each of their nodes contain a single digit.

上面的例子是：

2+5=7；

4+6=10；10不满足 a  single digit条件，0保存到当前结点中，进位1就要保存到下一个结点。

思路：

思路非常简单，分别遍历两个链表，并且用一个变量表示是否有进位。某个链表遍历结束之后再将另一个链表连接在结果链表之后即可，若最后有进位需要添加一位。

给出一种解答：Runtime: 420 ms

head 相当于头指针，方便遍历结果链表

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode result = new ListNode(0);
        ListNode head = result;
        int sum = 0;
        int carry = 0;//jin wei
        while(l1 != null || l2 != null) {
            int val1 = 0;
            if(l1 != null){
                val1 = l1.val;
                l1 = l1.next;
            }
            int val2 = 0;
            if(l2 != null) {
                val2 = l2.val;
                l2 = l2.next;
            }

            sum = val1 + val2 + carry;
            result.next = new ListNode(sum%10);
            result = result.next;
            carry = sum/10;

        }
        if(carry == 1){
            result.next = new ListNode(1);
        }
        return head.next;
    }
}

九章算法网给出的答案是：

Runtime: 540 ms

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if(l1 == null && l2 == null) {
            return null;
        }

        ListNode head = new ListNode(0);
        ListNode point = head;
        int carry = 0;
        while(l1 != null && l2!=null){
            int sum = carry + l1.val + l2.val;
            point.next = new ListNode(sum % 10);
            carry = sum / 10;
            l1 = l1.next;
            l2 = l2.next;
            point = point.next;
        }

        while(l1 != null) {
            int sum =  carry + l1.val;
            point.next = new ListNode(sum % 10);
            carry = sum /10;
            l1 = l1.next;
            point = point.next;
        }

        while(l2 != null) {
            int sum =  carry + l2.val;
            point.next = new ListNode(sum % 10);
            carry = sum /10;
            l2 = l2.next;
            point = point.next;
        }

        if (carry != 0) {
            point.next = new ListNode(carry);
        }
        return head.next;
    }
}