Add Two Numbers

Question:

You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

解法代码

public class LeetCode002 {

    public static void main(String[] args) {
        ListNode ln1 = new ListNode(2);
        ListNode ln2 = new ListNode(4);
        ListNode ln3 = new ListNode(3);
        ln1.next = ln2;
        ln2.next = ln3;
        
        ListNode ln21 = new ListNode(5);
        ListNode ln22 = new ListNode(6);
        ListNode ln23 = new ListNode(4);
        ln21.next = ln22;
        ln22.next = ln23;
        
        ListNode ln = addTwoNumbers(ln1, ln21);
        System.out.print(ln.val);
        while(ln.next != null) {
            System.out.print(ln.next.val);
            ln = ln.next;
        }
    }
    
    public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        
        ListNode re = new ListNode(0);//返回结果取re.next
        ListNode p = l1, q = l2, curNode = re;
        int carry = 0;//保存进位，值为0或1

        while(p != null || q != null) {
            int x = (p == null) ? 0 : p.val;//如果没有值则用0替代
            int y = (q == null) ? 0 : q.val;
            int sum = x + y + carry;
            
            carry = sum/10;
            curNode.next = new ListNode(sum%10);
            curNode = curNode.next;
            if(p != null) 
                p = p.next;
            if(q != null)
                q = q.next;
        }
        if(carry != 0) {//循环结束后，如果进位为1则增加此节点
            curNode.next = new ListNode(carry);
        }

        return re.next;//re头节点的下一个节点作为返回结果的头结点
    }
}

Output：

708

Time And Space Complexity

• addTwoNumbers

Time: 假设m,n表示l1,l2的长度，while循环最多执行max(m,n)次
Space: 新链表的长度最大为max(m,n)+2，一个是进位，一个是起始位置节点

Tips

注意以下测试用例