Leetcode:添加反向链接列表中表示的两个数字不起作用 [英] Leetcode: Adding two numbers represented in reverse linked list is not working

问题描述

以下代码不适用于以下输入:[2,4,3][5,6,4]输出:[7,8]预期的:[7,0,8]

The following code is not working for the below Input: [2,4,3] [5,6,4] Output: [7,8] Expected: [7,0,8]

为什么我没有得到0?谁能帮我.

Why I am not getting 0? Can anyone please help me.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode AddTwoNumbers(ListNode l1, ListNode l2) {
        return AddTwoNumbersHelper(l1, l2, 0);            
    }


private ListNode AddTwoNumbersHelper(ListNode l1, ListNode l2, int carry) {
    if (l1 == null && l2 == null)
        return null;

    int temp = 0;
    if (l1 != null)
        temp += l1.val;

    if (l2 != null)
        temp += l2.val;

    ListNode result = new ListNode(temp % 10);
    carry = temp / 10;

    l1 = l1.next;
    l2 = l2.next;
    int sum = 0;
    while(l1 != null || l2 != null) {
        sum = carry;

        if (l1 != null)
            sum += l1.val;

        if (l2 != null)
            sum += l2.val;

        carry = sum == 0 ? 0 : sum / 10;
        sum = sum % 10;

        result.next = new ListNode(sum);
        if(l1 != null)
            l1 = l1.next;
        if(l2 != null)
            l2 = l2.next;
    }

    if (carry > 0)
        result.next = new ListNode(carry);

    return result;
} 

}

推荐答案

代码存在一些缺陷:

1. 除非您要递归的不是递归参数，否则方法中不需要进位参数.
2. 列表提升步骤(l = l.next)可以与将节点的值添加到temp(sum)变量的步骤结合在一起.
3. 您不需要编写额外的代码来处理悬空的手提箱，循环就足够了，只需在循环中包含条件即可.
4. 该错误是因为您在每次迭代中不断更新头部的下一个指针，而不是在链接列表中构建(添加一个节点).您需要一个额外的指针/变量来做到这一点.
5. "sum"是比"temp"更好的变量名.

这是经过稍微修改的解决方案，可以解决上述问题:

Here's a slightly modified solution which fixes the above problems:

private ListNode AddTwoNumbersHelper(ListNode l1, ListNode l2) {

    ListNode result = null;
    ListNode tail = null;
    int carry = 0;
    while ((l1 != null) || (l2 != null) || (carry != 0)) {

        int sum = carry;
        if (l1 != null) {
            sum += l1.val;
            l1 = l1.next;
        }
        if (l2 != null) {
            sum += l2.val;
            l2 = l2.next;
        }
        carry = sum/10;
        sum %= 10;
        if (result == null) { // first time
            result = new ListNode(sum);
            tail = result;
        } else {
            tail.next = new ListNode(sum);
            tail = tail.next;
        }
    }
    return result;
}

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