题解 | 链表相加(二)

反转链表+进位模拟

由于加法需要从个位开始，所以最直观的策略就是将链表反转，从个位数开始，从左到右计算结果，计算完再反转回来即可

1. 反转链表：将 9->3->7 变为 7->3->9，将 6->3 变为 3->6
2. 同步遍历求和：从头开始相加，维护一个进位变量 carry
3. 构建新链表：每次计算出当前位的值（sum % 10）和进位值（sum / 10）
4. 最终反转：将得到的结果链表再次反转，恢复正确的高低位顺序

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 *   public ListNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param head1 ListNode类 
     * @param head2 ListNode类 
     * @return ListNode类
     */
    public ListNode addInList (ListNode head1, ListNode head2) {
        // write code here
        // 翻转链表
        head1 = reverse(head1);
        head2 = reverse(head2);

        ListNode dummy = new ListNode(-1);
        ListNode curr = dummy;
        int carry = 0;

        // 2.模拟加法逻辑
        while(head1 != null || head2 != null || carry != 0) {
            int val1 = (head1 != null) ? head1.val : 0;
            int val2 = (head2 != null) ? head2.val : 0;
            int sum = val1 + val2 + carry;
            carry = sum / 10;
            curr.next = new ListNode(sum % 10);
            curr = curr.next;

            if(head1 != null) head1 = head1.next;
            if(head2 != null) head2 = head2.next;
        }

        return reverse(dummy.next);
    }

    private ListNode reverse(ListNode head) {
        ListNode prev = null;
        ListNode curr = head;

        while(curr != null) {
            ListNode next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;
        }
        return prev;
    }
}