2021.03.23 打卡 LeetCode 2.两数相加

LeetCode 2.两数相加

时间复杂度O(n)

解法一：递归

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if (l2 == null) {
            return l1;
        } else if (l1 == null && l2 != null) {
            return l2;
        }else {
            l1.val = l1.val + l2.val;
            jinwei(l1);
            l1.next = addTwoNumbers(l1.next, l2.next);
            return l1;
        }
    }

    public void jinwei(ListNode l1) {
        if (l1.val > 9){
            if (l1.next == null){
                l1.next = new ListNode(l1.val/10);
            }else {
                l1.next.val += l1.val / 10;
                jinwei(l1.next);
            }
            l1.val %= 10;
        }
    }
}

//runtime:2 ms
//memory:39.3 MB

解法二：链表计数，如果大于10就进一位

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        int i = 0;
        ListNode root = new ListNode(0);
        ListNode node = root;
        while(l1!= null || l2 !=null || i != 0){
            int v1 = l1 != null? l1.val: 0;
            int v2 = l2 != null? l2.val: 0;

            ListNode sNode = new ListNode((v1 + v2 + i) % 10);
            node.next = sNode;
            node = sNode;

            i = (v1 + v2 + i) / 10;
            if(l1 != null) l1 = l1.next;
            if(l2 != null) l2 = l2.next;
        }
        return root.next;
    }
}
//runtime:2 ms
//memory:39 MB