LeetCode Add Two Numbers（002）解法总结

描述

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.

思路

使用int类型的加法实现，先转成加法再进行分割成链表

/**
 * 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) {
        int i1 = 0, i2 = 0, i3 = 0, base = 1;
        ListNode temp = l1;
        while(temp!=null){
            i1 = i1 + temp.val*base;
            temp = temp.next;
            base = base * 10;
        }
        temp = l2;
        base = 1;
        while(temp!=null){
            i2 = i2 + temp.val*base;
            temp = temp.next;
            base = base * 10;
        }
        i3 = i1 + i2;
        String out = Integer.toString(i3);
        
        temp = null;
        for(int i = 0 ;i <out.length();i++){
            int num = (int)out.charAt(i) - (int)('0');
            ListNode newNode = new ListNode(num);
            newNode.next = temp;
            temp = newNode;
        }
        return temp;
    }
}

结果发生了溢出

变换思路

使用按位相加、进位的方法生成链表

一开始想着进位，忽视进位的情况不止一种，改为用int进行存储，还能累加进位。

使用两个指针保证从链表头进行输出。

/**
 * 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 tail = head;
        int carry = 0;
        while(l1 != null || l2 != null || carry != 0){
            if(l1 != null){
                carry = carry + l1.val;
                l1 = l1.next;
            }
            if(l2 != null){
                carry = carry + l2.val;
                l2 = l2.next;
            }
            tail.next = new ListNode(carry % 10);
            carry = carry / 10;
            tail = tail.next;
        }
        return head.next;
    }
}