[leetcode][2] Add Two Numbers

标签：repr   list   sel   ext   node   dig   leading   leetcode   简单   

2. Add Two Numbers

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.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

解析

很简单用链表表示数字相加，注意进位就行了

参考答案（自己写的）

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode head = new ListNode(0);
        ListNode index = head;
        int move = 0;
        while(l1 != null || l2 != null) {
            int value = move;
            if (l1 != null) {
                value += l1.val;
                l1 = l1.next;
            }
            if (l2 != null) {
                value += l2.val;
                l2 = l2.next;
            }
            if (value >= 10) {
                move = 1;
                value = value - 10;
            } else {
                move = 0;
            }
            ListNode newNode = new ListNode(value);
            index.next = newNode;
            index = newNode;
        }
        
        if (move != 0) {
            index.next = new ListNode(1);
        }
        
        return head.next;
    }
}

注意点原来我用的是value = value%10,这样效率很低，其实value的值不会超过10，所以直接value = value - 10就行了。

[leetcode][2] Add Two Numbers

标签：repr   list   sel   ext   node   dig   leading   leetcode   简单   

原文地址：https://www.cnblogs.com/ekoeko/p/9604875.html