首先第一种方法 两个单个节点还有进位相加实现如下: 1/**2* Definition for singly-linked list.3* public class ListNode {4* int val;5* ListNode next;6* ListNode(int x) { val = x; }7* }8*/9publicclassSolution {10publicListNode addTwoNumbers(ListNode l1, ListNode l2) {11if(l1==null&&...
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, exc...
下面是将list转化为linked lists的函数: Figure 2-2 这样,list里的每一位数都储存在link lists里了,增加的linked lists的长度就是list的长度。这里dummy root是构造linked lists常用的一种形式,将dummy root作为一个参照,每次指向下一个node从而按要求构造好linked lists, 而dummy node就相当于构造的链表的头结点。
两数相加(addTwoNumbers) 两数之和是linked list中常见的问题之一,经常用于检测coder是否理解linked list和对linked list的操作。 问题一般是:给定两个非空的linked list,表示两个非负整数,每一个结点包含一个数字digit。这些数字digits都是方向存储的。将这两个数相加之后返回总和且同样地以反向储存的方式储存在link...
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 Explanation: 342 + 465 = 807. 思路: 小学加法思路,判断是否需要进一,用left来做记录。用dummy来记录。(dummy其实只是一个nod...
You are given twonon-emptylinked lists representing two non-negative integers. The digits are stored inreverse orderand 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 th...
2. Add Two Numbers(链表尾插法) You are given twonon-emptylinked lists representing two non-negative integers. The digits are stored inreverse orderand each of their nodes contain a single digit. Add the two numbers and return it as a linked list....
解法2 遍历两个链表,把各个位数相加,注意进位 代码语言:javascript 复制 publicstaticListNodeaddTwoNumbers2(ListNode l1,ListNode l2){// 边界条件判断if(l1==null){returnl2;}elseif(l2==null){returnl1;}ListNode list=null;ListNode next=null;// 记录和值int sum=0;// 记录是否有进位int b=0;while(l1...
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, exc...
3 输入与输出：/*** Definition for singly-linked list.* struct ListNode {* int val;* ListNode *next;* ListNode(int x) : val(x), next(NULL) {}* };*/class Solution {public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { }};4 解决思路：从表头开始相加，记录每次相加...