2. Add Two Numbers

最新推荐文章于 2024-01-03 15:29:28 发布

dihu6677

最新推荐文章于 2024-01-03 15:29:28 发布

阅读量115

点赞数

原文链接：http://www.cnblogs.com/GodBug/p/8490820.html

版权

本文介绍了一种算法，用于解决两个非空链表表示的非负整数相加的问题。链表中的数字以逆序存储，每个节点包含一个数字。通过迭代处理每个节点的值，并考虑进位，最终将结果返回为一个新的链表。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

/**
 * 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) {
        // save result node
        ListNode resNode = new ListNode(0);
        // carry bit flag
        boolean isCarryBit = false;
        // position flag
        boolean isFirstIndex = true;
        // temp node
        ListNode tempNode = resNode;
        // iterator node
        while(l1 != null || l2 != null) {
            int val1 = l1 != null ? l1.val : 0;
            int val2 = l2 != null ? l2.val : 0;
            // sum digit
            int sum = val1 + val2;
            int currentVal = (sum + (isCarryBit ? 1 : 0)) % 10;
            // crray bit handle
            isCarryBit = ((sum + (isCarryBit ? 1 : 0)) >= 10 ? true : false);
            if(isFirstIndex) {
                resNode.val = currentVal;
            } else {
                tempNode.next = new ListNode(currentVal);
                tempNode = tempNode.next;
            }
            isFirstIndex = false;
            // evalution 
            if(l1 != null){
                l1 = l1.next;
            }
            if(l2 != null){
                l2 = l2.next;
            }
        }
        // carry bit
        if(isCarryBit){
            tempNode.next = new ListNode(1);
        }
        // return result node
        return resNode;
    }
}

转载于:https://www.cnblogs.com/GodBug/p/8490820.html