LeetCode: Add Two Numbers

文章目录

You are given two linked lists representing two non-negative numbers. 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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)

Output: 7 -> 0 -> 8

很基础的链表操作题目，一次循环依次对节点进行加法运算即可，要注意进位与边界条件，时间复杂度O(n)

为了可理解性，在我的实现中把加法与进位运算拆开分别处理了：

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    if (l1 == null || l2 == null) {
        return l1 == null ? l2 : l1;
    }

    ListNode p = new ListNode(l1.val + l2.val);
    ListNode head = p;
    while (l1.next != null && l2.next != null) { // 加法操作
        l1 = l1.next;
        l2 = l2.next;
        p.next = new ListNode(l1.val + l2.val);
        p = p.next;
    }
    l1 = l1.next;
    l2 = l2.next;
    p.next = l1 == null ? l2 : l1;

    p = head;
    while (p != null) { // 处理进位
        if (p.val > 9) {
            if (p.next == null) {
                p.next = new ListNode(p.val / 10);
            } else {
                p.next.val += p.val / 10;
            }
            p.val %= 10;
        }
        p = p.next;
    }
    return head;
}