即实现342+465=807 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) ...
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...
代码: defaddTwoNumbers(self,l1,l2):""" :type l1: ListNode :type l2: ListNode :rtype: ListNode """curr=dummy=ListNode(0)left=0whilel1orl2orleft:(l1,num1)=(l1.next,l1.val)ifl1else(None,0)(l2,num2)=(l2.next,l2.val)ifl2else(None,0)left,num=divmod(num1+num2+left,10)curr...
Leetcode 刷题指北 2. Add Two Numbers You are given twonon-emptylinked lists representing twonon-negativeintegers. 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 no...
* Definition for singly-linked list. * class ListNode{* int val; * ListNode next; * ListNode(){}* ListNode(int val){this.val = val;}* ListNode(int val, ListNode next){this.val = val; this.next = next;}*}*/publicListNodeaddTwoNumbers(ListNodel1,ListNodel2){ListNodedummy=newListNode(...
addTwoNumbers(l1, l2); 72 while (l != NULL){ 73 cout << l->val << endl; 74 l = l->next; 75 } 76 while (1); 77 } 运行结果: 最后实现不等长有进位的求和,即实现题目要求(注意最后一位有进位的情况) 代码语言:javascript 复制 1 #include <iostream> 2 3 using namespace std; 4 ...
The key of this problem is to think of using Stack 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) {11Stack<ListN...
order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list. Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8 分析:342+465=807——>708 Add.h #pragmaonce#include<iostream>usingnamespacestd;typedefstructListNode{int_...
Add the two numbers and return it as a linked list. Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8 分析:342+465=807——>708 Add.h #pragma once #include using namespace std; typedef struct ListNode{ int _var; ListNode *_next; ListNode(int var) :_var(...
order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list. Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8 分析:342+465=807——>708 Add.h #pragmaonce#include<iostream>usingnamespacestd;typedefstructListNode{int_...