leetCode链表加和

1、源码，转自：https://blog.youkuaiyun.com/sinat_41144773/article/details/88091467

1）

public class ListNode {
    int val;
    ListNode next;   // 下一个链表对象
    ListNode(int x) { val = x; }  //赋值链表的值

2）

package test;

import org.testng.annotations.Test;

public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
//生成ListNode链表对象，链表的值为0，没有指向的节点
        ListNode dummyHead = new ListNode(0);
        ListNode p = l1, q = l2, curr = dummyHead;
        int carry = 0;
        while (p != null || q != null) {

//两个链表的同一位赋值给变量x和y
            int x = (p != null) ? p.val : 0;
            int y = (q != null) ? q.val : 0;


            int sum = carry + x + y;
            carry = sum / 10; //这里的carry循环时在上面这个式子用int sum = carry + x + y;


            curr.next = new ListNode(sum % 10);  //如果结果是两位数，各位数留在结果链表里
            curr = curr.next;


            if (p != null) p = p.next;
            if (q != null) q = q.next;
        }

//最后一位的进位，如果有进位，把进位放到下一个链表里
        if (carry > 0) {
            curr.next = new ListNode(carry);
        }
        return dummyHead.next;

    }
    @Test
    public void test()
    {
        ListNode L1=new ListNode(3);
        ListNode L1_2=new ListNode(4);
        ListNode L1_3=new ListNode(2);
        L1.next=L1_2;
        L1_2.next=L1_3;
        L1_3.next=null;

        ListNode L2=new ListNode(4);
        ListNode L2_2=new ListNode(6);
        ListNode L2_3=new ListNode(5);
        L2.next=L2_2;
        L2_2.next=L2_3;
        L2_3.next=null;
        ListNode solution=addTwoNumbers(L1,L2);

        while(solution!=null)
        {
            System.out.println(solution.val);
            solution=solution.next;

        }


    }
}

 

 

2、运行结果

7

0

8