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#yyds干货盘点# LeetCode 热题 HOT 100：两数相加
source link: https://blog.51cto.com/u_13321676/5662152
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#yyds干货盘点# LeetCode 热题 HOT 100：两数相加
精选 原创给你两个 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 的方式存储的，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。
输入：l1 = [2,4,3], l2 = [5,6,4]
输出：[7,0,8]
解释：342 + 465 = 807.
输入：l1 = [0], l2 = [0]
输出：[0]
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]
代码实现：
/**
* 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 head = new ListNode(-1);
ListNode cur = head;
int count = 0;
while(l1 != null || l2 != null){
int v1 = 0;
if(l1 != null){
v1 = l1.val;
l1 = l1.next;
}
int v2 = 0;
if(l2 != null){
v2 = l2.val;
l2 = l2.next;
}
int sum = v1 + v2 + count;
cur.next = new ListNode(sum % 10);
cur = cur.next;
count = sum / 10;
}
if(count > 0){
cur.next = new ListNode(count);
}
return head.next;
}
}
* 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 head = new ListNode(-1);
ListNode cur = head;
int count = 0;
while(l1 != null || l2 != null){
int v1 = 0;
if(l1 != null){
v1 = l1.val;
l1 = l1.next;
}
int v2 = 0;
if(l2 != null){
v2 = l2.val;
l2 = l2.next;
}
int sum = v1 + v2 + count;
cur.next = new ListNode(sum % 10);
cur = cur.next;
count = sum / 10;
}
if(count > 0){
cur.next = new ListNode(count);
}
return head.next;
}
}
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