LeetCode 2 Add Two Numbers

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.

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
	*/
class Solution {
	public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    	ListNode list = new ListNode(0);
    	ListNode node = list;
    	int count = 0;
    	while(l1 != null && l2 != null){
        	//count = (l1.val + l2.val) / 10;
        	int sum = l1.val + l2.val + count;
        	count = sum / 10;
        	node.next = new ListNode(sum % 10);
        	node = node.next;
        	l1 = l1.next;
        	l2 = l2.next;
   		 }
    	while(l1 != null){
        	int sum = l1.val + count;
        	count = sum / 10;
        	node.next = new ListNode(sum % 10);
        	node = node.next;
        	l1 = l1.next;
    	}
    	while(l2 != null){
        	int sum = l2.val + count;
        	count = sum / 10;
        	node.next = new ListNode(sum % 10);
        	node = node.next;
        	l2 = l2.next;
    	}
    	if(count > 0){
        	node.next = new ListNode(count);
    	}
    	return list.next;
	}
}

官方的題解：

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
	ListNode dummyHead = new ListNode(0);
	ListNode p = l1, q = l2, curr = dummyHead;
	int carry = 0;
	while (p != null || q != null) {
   	 	int x = (p != null) ? p.val : 0;
    	int y = (q != null) ? q.val : 0;
    	int sum = carry + x + y;
    	carry = sum / 10;
    	curr.next = new ListNode(sum % 10);
    	curr = curr.next;
    	if (p != null) p = p.next;
    	if (q != null) q = q.next;
	}
	if (carry > 0) {
    	curr.next = new ListNode(carry);
	}
	return dummyHead.next;
}

時間複雜度：O（n）
空間複雜度：O（n）