leetcode: #2 两数相加

leetcode: #2 两数相加

题目描述

java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 pointer1 = l1;
        ListNode pointer2 = l2;
        int sumOfRemainder = 0;
        int carryOfNumber = 0;
        ListNode headNode = new ListNode();
        ListNode currentNode = headNode;

        while(pointer1 != null && pointer2 !=null) {
    
    
            sumOfRemainder = (carryOfNumber + pointer1.val + pointer2.val) % 10;
            carryOfNumber = (carryOfNumber + pointer1.val + pointer2.val) / 10;
            ListNode tempNode = new ListNode(sumOfRemainder);
            currentNode.next = tempNode;
            currentNode = tempNode;
            
            pointer1 = pointer1.next;
            pointer2 = pointer2.next;
        }

        while(pointer1 != null) {
    
    
            sumOfRemainder = (carryOfNumber + pointer1.val) % 10;
            carryOfNumber = (carryOfNumber + pointer1.val) / 10;
            ListNode tempNode = new ListNode(sumOfRemainder);
            currentNode.next = tempNode;
            currentNode = tempNode;

            pointer1 = pointer1.next;
        }

        while(pointer2 != null) {
    
    
            sumOfRemainder = (carryOfNumber + pointer2.val) % 10;
            carryOfNumber = (carryOfNumber + pointer2.val) / 10;
            ListNode tempNode = new ListNode(sumOfRemainder);
            currentNode.next = tempNode;
            currentNode = tempNode;
            
            pointer2 = pointer2.next;
        }

        if(carryOfNumber != 0) {
    
    
            ListNode tempNode = new ListNode(carryOfNumber);
            currentNode.next = tempNode;
        }

        return headNode.next;
    }
}