【反转链表】力扣 445. 两数相加 II

一、题目

二、思路

• 加法运算是从低位开始，向高位进位，因此需要将两个链表进行反转，再进行对齐后的相加操作。
• 力扣 2. 两数相加

三、题解

/**
 * 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 ans;
        l1 = reverseList(l1);
        l2 = reverseList(l2);
       return reverseList(addTwo(l1, l2, 0));
    }
    // 反转链表
    public ListNode reverseList(ListNode head) {
        ListNode pre = null, cur = head;
        while (cur != null) {
            ListNode tmp = cur.next;
            cur.next = pre;
            pre = cur;
            cur = tmp;
        }
        return pre;
    }
    // 求解两数之和 
    public ListNode addTwo(ListNode l1, ListNode l2, int carry) {
        // 递归出口
        if (l1 == null && l2 == null) {
            return carry != 0 ? new ListNode(carry) : null;
        }
        // 确保 l1 是较长的
        if (l1 == null) {
            l1 = l2;
            l2 = null;
        }
        int sum = l1.val + (l2 != null ? l2.val : 0) + carry;
        l1.val = sum % 10;
        l1.next = addTwo(l1.next, l2 != null ? l2.next : null, sum / 10);
        return l1;
    }
}