算法#25--两整数链表求和_tcl求和0.1+0.2+0.3+......+n-优快云博客

本文链接：https://blog.youkuaiyun.com/tclxspy/article/details/53389062

本文介绍了一种算法挑战，即如何将两个表示为逆序链表形式的非负整数相加，并返回结果作为新的链表。提供了两种解决方案：一种较为直观的方法和一种更为简洁高效的实现方式。

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

My Code:


/**
 * 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) {
        ListNode first = null;
        ListNode last = null;
        int v = 0;
        while (true)
        {
            ListNode oldlast = last;
            if (l1 == null && l2 == null && v == 0)
            {
                break;
            }

            if (l1 == null)
            {
                l1 = new ListNode(0);
            }

            if (l2 == null)
            {
                l2 = new ListNode(0);
            }

            int s = l1.val + l2.val + v;
            v = s / 10;
            last = new ListNode(s % 10);
            l1 = l1.next;
            l2 = l2.next;                

            if (first == null)
            {
                first = last;
            }
            else
            {
                oldlast.next = last;
            } 
        }
        return first;
    }
}

Best Code:


public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode c1 = l1;
        ListNode c2 = l2;
        ListNode sentinel = new ListNode(0);
        ListNode d = sentinel;
        int sum = 0;
        while (c1 != null || c2 != null) {
            sum /= 10;
            if (c1 != null) {
                sum += c1.val;
                c1 = c1.next;
            }
            if (c2 != null) {
                sum += c2.val;
                c2 = c2.next;
            }
            d.next = new ListNode(sum % 10);
            d = d.next;
        }
        if (sum / 10 == 1)
            d.next = new ListNode(1);
        return sentinel.next;
    }
}