力扣2-两数之和

 
 namespace Test2;
 
 // Definition for singly-linked list.
 public class ListNode {
      public int val;
      public ListNode next;
      public ListNode(int val=0, ListNode next=null) {
          this.val = val;
          this.next = next;
      }
  }
 
public class Solution3 {

    private ListNode CreateListNodeByArr(int[] arr,int index)
    {
        if(index == arr.Length - 1) return null;
        var root = new ListNode(arr[index]);
        root.next = CreateListNodeByArr(arr, index + 1);
        return root;
    }
    public void Show()
    {
        int[] arr1 = {9,9,9,9,9,9,9};
        int[] arr2 = {9,9,9,9};
        // var l1 = CreateListNodeByArr(arr1,0);
        // var l2 = CreateListNodeByArr(arr2,0);
        // 链表构造的方法。。。。。。。
        ListNode l1 = new ListNode(9);
        ListNode l1Cur = l1;
        foreach(var item in arr1)
        {
            l1Cur.next = new ListNode(item);
            l1Cur = l1Cur.next;
        }
        l1 = l1.next;

        ListNode l2 = new ListNode(9);
        ListNode l2Cur = l2;
        foreach(var item in arr2)
        {
            l2Cur.next = new ListNode(item);
            l2Cur = l2Cur.next;
        }
        l2 = l2.next;
        

        // l1.next = new ListNode(9);
        // l1.next.next = new ListNode(3);
        // ListNode l2 = new ListNode(5);
        // l2.next = new ListNode(6);
        // l2.next.next = new ListNode(4);
        var rt = AddTwoNumbers(l1, l2);
        Console.WriteLine(rt.val);

    }

    public ListNode AddTwoNumbers(ListNode l1, ListNode l2) {
        // 递推
        ListNode pre = new ListNode();
        ListNode cur = pre; // 这里的没看懂,这里使用就不会改变pre的值。
        
        var carry = (l1.val + l2.val) / 10;
        while(l1 != null || l2 != null || carry != 0)
        {
            var x = l1 == null ? 0 : l1.val;
            var y = l2 == null ? 0 : l2.val;
            var val = (x + y + carry) % 10;
            cur.next = new ListNode(val);
            cur = cur.next;
            carry = (x + y + carry) / 10;
            l1 = l1 == null ? null : l1.next;
            l2 = l2 == null ? null : l2.next;
        }
    
        return pre.next;
    }
}