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. 翻译 给定两个非空的链表,代表两个非负整数。这两个整数都是倒叙存储,要求返回一个链表,表示这两个整数的和。 样例 Input: (2 -> 4 -> 3) +...
next(NULL) {}* };*/classSolution{public:ListNode*addTwoNumbers(ListNode* l1, ListNode* l2){ListNode * p1 = l1;ListNode * p2 = l2;ListNode * p3 =NULL;ListNode * p = p3;intin =0;//遍历对应的位置while(p1 && p2){intsum = p1->val ...
https://leetcode-cn.com/problems/add-two-numbers/description/ 给定两个非空链表来表示两个非负整数。位数按照逆序方式存储,它们的每个节点只存储单个数字。将两数相加返回一个新的链表。 解法1 两个链表长度相同的部分,对位相加,注意需要还需要加上上一次相加的进位值。得到结果后,将个位数存入结果链表中,十...
* type ListNode struct { * Val int * Next *ListNode * } */ func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode { // 哨兵结点,方便后续处理 head_pre := &ListNode{} // 结果链表的尾结点,方便用尾插法插入 tail := head_pre // 进位值,初始化为 0 carry := 0 // 如果两个链表...
LeetCode 2 Add Two Numbers——用链表模拟加法 上方蓝字,和我一起学技术。 今天要讲的是一道经典的算法题,虽然不难,但是很有意思,我们一起来看下题目: 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 ...
LeetCode: 2. Add Two Numbers LeetCode: 2. Add Two Numbers 题目描述 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...
def addTwoNumbers(self, l1, l2): “”" :type l1: ListNode :type l2: ListNode :rtype: ListNode “”" p = 0 result = None current = None while True: addValue = l1.val + l2.val + p p = 0 if addValue >= 10: addValue = addValue - 10 ...
public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode result = new ListNode(0); ListNode tem = l1; ListNode tem2 = l2; ListNode current = result; boolean flag = false; while (tem != null || tem2 != null) { // 计算节点值之和,需要判空 int sum = (tem != null ?
Add Two Numbers 二、解题 1)题意 给出两个链表,把对应位置上的值进行十进制相加(有进位),返回链表的根节点。 2)输入输出说明 输入:两个列表的根节点(并不是整个列表,即leetcode会把默认生成好的列表的根节点传入) 输出:累加之后的根节点 3)关键点 ...
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!=null||l2!=null){if(l1!=null){sum=l1.val;l1=l1.next;}if...