2. Add Two Numbers

最新推荐文章于 2024-01-03 15:29:28 发布

转载最新推荐文章于 2024-01-03 15:29:28 发布 · 43 阅读

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原文链接：http://www.cnblogs.com/YaungOu/p/6718714.html

本文介绍了一种解决两数相加问题的方法，通过链表形式存储逆序的数字，实现两个数字的相加并返回结果链表。算法的时间复杂度为O(n+m)，空间复杂度为O(1)。

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题目：

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.

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

复杂度：

时间：O(n+m)、空间：O(1)

实现：

/**
 * 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 headNode = null, preNode = null;
        int carry = 0;
        while(l1 != null || l2 != null || carry > 0) {
            int sum = (l1 != null ? l1.val : 0) + (l2 != null ? l2.val : 0) + carry;
            carry = sum / 10;
            
            ListNode curNode = new ListNode(sum % 10);
            if(preNode == null) {
                headNode = curNode;
            } else {
                preNode.next = curNode;
            }
            preNode  = curNode;
            
            if(l1 != null && l1.next != null) {
                l1 = l1.next;
            } else {
                l1 = null;
            }
            
            if(l2 != null && l2.next != null) {
                l2 = l2.next;
            } else {
                l2 = null;
            }
        }
        
        return headNode;
    }
}

转载于:https://www.cnblogs.com/YaungOu/p/6718714.html