Leetcode Add Two Numbers & 两数相加 解题报告

1.问题描述

给你两个非空的链表,表示两个非负的整数。它们每位数字都是按照逆序方式存储的,并且每个节点只能存储一位数字。
请你将两个数相加,并以相同形式返回一个表示和的链表。
你可以假设除了数字 0 之外,这两个数都不会以 0 开头。

2.测试用例

示例 1:
输入:l1 = [2,4,3], l2 = [5,6,4]
输出:[7,0,8]
解释:342 + 465 = 807.

示例 2:
输入:l1 = [0], l2 = [0]
输出:[0]

示例 3:
输入:l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
输出:[8,9,9,9,0,0,0,1]

3.补充说明

每个链表中的节点数在范围 [1, 100] 内
0 <= Node.val <= 9
题目数据保证列表表示的数字不含前导零

4.解题报告

  1. 递归
    /**
     * Definition for singly-linked list.
     * public class ListNode {
     *     int val;
     *     ListNode next;
     *     ListNode() {}
     *     ListNode(int val) { this.val = val; }
     *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
     * }
     */
    class Solution {
        
        public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
            ListNode currentNode = new ListNode();
            int next = 0;
            currentNode = assignListNode(l1, l2, currentNode, next);
            return currentNode;
        }
        
        public ListNode assignListNode(ListNode l1, ListNode l2, ListNode currentNode, int next) {
            if (l1 == null && l2 == null) {
                if (next > 0) {
                    currentNode.val = 1;
                } else {
                    currentNode = null;
                }
                return currentNode;
            }
            int x = l1 == null ? 0 : l1.val;
            int y = l2 == null ? 0 : l2.val;
            int sum = x + y + next;
            currentNode.val = sum % 10;
            currentNode.next = new ListNode();
            currentNode.next = assignListNode(l1 == null ? null : l1.next, l2 == null ? null : l2.next, currentNode.next, sum / 10);
            return currentNode;
        }
    }
    
  1. 内循环
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode dummyHead = new ListNode(0);
        ListNode p = l1, q = l2, curr = dummyHead;
        int carry = 0;
        while (p != null || q != null) {
            int x = (p != null) ? p.val : 0;
            int y = (q != null) ? q.val : 0;
            int sum = carry + x + y;
            carry = sum / 10;
            curr.next = new ListNode(sum % 10);
            curr = curr.next;
            if (p != null) p = p.next;
            if (q != null) q = q.next;
        }
        if (carry > 0) {
            curr.next = new ListNode(carry);
        }
        return dummyHead.next;
    }
    

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