leetcode : Add two numbers 解题报告

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

 

Tag : dummy node

 

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        
        if(l1 == null && l2 == null) {
            return null;
        }
        
        if(l1 == null) {
            return l2;
        }
        
        if(l2 == null) {
            return l1;
        }
        
        ListNode head = new ListNode(0);
        ListNode dummy = head;
        int carry = 0;
        
        while(l1 != null && l2 != null) {
            int val = (l1.val + l2.val + carry) % 10;
            carry = (l1.val + l2.val + carry) / 10;
            ListNode node = new ListNode(val);
            head.next = node;
            head = node;
            l1 = l1.next;
            l2 = l2.next;
        }
        while(l1 != null) {
            int val = (l1.val + carry) % 10;
            carry = (l1.val + carry) / 10;
            ListNode node = new ListNode(val);
            head.next = node;
            head = node;
            l1 = l1.next;
        }
        while(l2 != null) {
            int val = (l2.val + carry) % 10;
            carry = (l2.val + carry) / 10;
            ListNode node = new ListNode(val);
            head.next = node;
            head = node;
            l2 = l2.next;
        }
        if(carry != 0) {
            ListNode node = new ListNode(carry);
            head.next = node;
        }
        return dummy.next;
    }
}

leetcode 最佳参考解法:

public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode c1 = l1;
        ListNode c2 = l2;
        ListNode sentinel = new ListNode(0);
        ListNode d = sentinel;
        int sum = 0;
        while (c1 != null || c2 != null) {
            sum /= 10;
            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)
            d.next = new ListNode(1);
        return sentinel.next;
    }
}
原文地址:https://www.cnblogs.com/superzhaochao/p/6358608.html