leetcode(2):Add Two Numbers

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

大致的意思是两个列表相加 2+5 4+6 3+4 至于最后是 0 8 因为加起来是 7 10 7 （10 进位 成为7 0 8） （注意进位）

源代码

package AddTwoNumbers;

public class Main {
public static void main(String args[]) {
ListNode l=new ListNode(2);
ListNode h=new ListNode(5);
ListNode l1=new ListNode(4);
ListNode l2=new ListNode(3);
ListNode l3=new ListNode(6);
ListNode l4=new ListNode(4);
l.next=l1;
l1.next=l2;
h.next=l3;
l3.next=l4;
print(  addTwoNumbers(l,h));
}
public static ListNode addTwoNumbers(ListNode c1, ListNode c2) {
ListNode sen = new ListNode(0);//  保存 生成  新的列表的  根节点引用
ListNode d = sen;
int sum = 0;
while (c1 != null || c2 != null) {
sum /= 10;//进位处理
if (c1 != null) {
sum += c1.val;
c1 = c1.next;
}
if (c2 != null) {
sum += c2.val;
c2 = c2.next;
}
d.next = new ListNode(sum % 10);
d = d.next;
}
if (sum / 10 >= 1)//处理最后遗留的进位
d.next = new ListNode(sum / 10);
return sen.next; //当前的根节点  是空的   在上文中创建并没有赋值   真正的根节点是他的下一个节点
}
public static void print(ListNode c1) {
while (c1 != null)
{
System.out.print(c1.val);
if(c1.next!=null)
{
System.out.print("->");
}
c1=c1.next;

}
}
}
class ListNode {
int val;
ListNode next;
ListNode(int x) { val = x; }
}

文章地址：http://www.haha174.top/article/details/256262

源码地址：https://github.com/haha174/daylx.git