LeetCode 2. 两数相加

2. 两数相加

Solution

思路：模拟。自己的写法是两个一起循环，其中一个结束就停止循环，再处理另一个链表，不过也可以一直循环，但是可以直接连接过去，然后就是考虑进位。

//class ListNode {
//  int val;
//  ListNode next;
//  ListNode() {}
//  ListNode(int val) { this.val = val; }
//  ListNode(int val, ListNode next) { this.val = val; this.next = next; }
//}

//写法有点点丑
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode root = new ListNode();
        int preDigit = 0;
        ListNode p = root;
        while (l1 != null && l2 != null) {
            ListNode t = new ListNode();
            int x = l1.val, y = l2.val;
            t.val = (x + y + preDigit) % 10;
            preDigit = (x + y + preDigit) / 10;
            p.next = t;
            p = t;
            l1 = l1.next;
            l2 = l2.next;
        }
        if (l2 != null) {
            l1 = l2;
        }
        while (l1 != null) {
            ListNode t = new ListNode();
            t.val = (l1.val + preDigit) % 10;
            preDigit = (l1.val + preDigit) / 10;
            if (preDigit == 0) {
                t.next = l1.next;
                p.next = t;
                break;
            }
            p.next = t;
            p = t;
            l1 = l1.next;
        }
        if (preDigit != 0) {
            ListNode t = new ListNode();
            t.val = preDigit;
            p.next = t;
            t.next = null;
        }
        return root.next;
    }
}


// 优雅 但是内存多一点点
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode head = null, tail = null;
        int digit = 0;
        while (l1 != null || l2 != null) {
            int val1 = l1 != null ? l1.val : 0;
            int val2 = l2 != null ? l2.val : 0;
            int sum = val1 + val2 + digit;
            if (head == null) {
                head = new ListNode(sum % 10);
                tail = head;
            } else {
                tail.next = new ListNode(sum % 10);
                tail = tail.next;
            }
            digit = sum / 10;
            if (l1 != null)
                l1 = l1.next;
            if (l2 != null)
                l2 = l2.next;
        }
        if (digit != 0) {
            tail.next = new ListNode(digit);
        }
        return head;
    }
}