34. Add two numbers

2. Addition of two numbers

Here are two for you Non empty The linked list of , Represents two nonnegative integers . Each of them is based on The reverse Stored in , And each node can only store a Numbers .

Please add up the two numbers , And returns a linked list representing sum in the same form .

You can assume that in addition to the numbers 0 outside , Neither of these numbers 0 start .

Example 1：

 Input ：l1 = [2,4,3], l2 = [5,6,4]
 Output ：[7,0,8]
 explain ：342 + 465 = 807.

Example 2：

 Input ：l1 = [0], l2 = [0]
 Output ：[0]

Example 3：

 Input ：l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
 Output ：[8,9,9,9,0,0,0,1]

Just go through it

1、 At the same time, enumerate two linked lists from scratch , take l1 and l2 The element pointed to by the pointer is saved to t in , then t % 10 The element of is saved to dummy In the list , Again t / 10 Remove the stored elements ,l1 and l2 At the same time, move back one space

2、 When all elements are traversed , If t != 0, And then t Deposit to dummy In the list

/** * 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 dummy = new ListNode(0);
    ListNode p = dummy;
    int t = 0;
    while(l1!=null||l2!=null){
    
      if(l1!=null) {
    t = t+l1.val;l1=l1.next;}
      if(l2!=null) {
    t = t +l2.val;l2 = l2.next;}
      ListNode a = new ListNode(t%10);
      p.next = a;
      p = a;
      t = t/10;
    }
    if(t!=0){
    
      ListNode a = new ListNode(t%10);
      p.next = a;
      p = a;
    }

    return dummy.next;

  }
}