标签：Medium ListNode sum Two next Add l2 l1 carry

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.

Related Topics: Linked List, Math

解题思路：由于代表两个数的链表已经逆序，从个位开始计算，所以结果中每一位是由对应位置的两个数相加，以及前一位的进位组成的。结果链表的最前面设一个dummy node，便于最后返回结果的处理，直接返回dummy.next。

Time Complexity: O(n), Space Complexity: O(1)

Java version:

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) { val = x; }
 7  * }
 8  */
 9 class Solution {
10     public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
11         ListNode dummy = new ListNode(0);
12         ListNode result = dummy;
13         
14         int carry = 0;
15         
16         while (l1 != null || l2 != null || carry == 1) {
17             int sum = carry;
18             
19             if (l1 != null) {
20                 sum += l1.val;
21                 l1 = l1.next;
22             }
23             if (l2 != null) {
24                 sum += l2.val;
25                 l2 = l2.next;
26             }
27             
28             ListNode curr = new ListNode(sum % 10);
29             carry = sum / 10;
30             
31             result.next = curr;
32             result = result.next;
33         }
34         
35         return dummy.next;
36     }
37 }

JavaScript version:

 1 /**
 2  * Definition for singly-linked list.
 3  * function ListNode(val) {
 4  *     this.val = val;
 5  *     this.next = null;
 6  * }
 7  */
 8 /**
 9  * @param {ListNode} l1
10  * @param {ListNode} l2
11  * @return {ListNode}
12  */
13 var addTwoNumbers = function(l1, l2) {
14     var dummy = new ListNode(0);
15     var head = dummy;
16     
17     var carry = 0;
18     
19     while (l1 != null || l2 != null || carry == 1) {
20         var sum = carry;
21         
22         if (l1 != null) {
23             sum += l1.val;
24             l1 = l1.next;
25         }
26         if (l2 != null) {
27             sum += l2.val;
28             l2 = l2.next;
29         }
30         // 注意这里javascript除法会产生float
31         carry = Math.floor(sum / 10);
32         var curr = new ListNode(sum % 10);
33         head.next = curr;
34         head = head.next;
35     }
36     
37     return dummy.next;
38 };

Python version:

 1 # Definition for singly-linked list.
 2 # class ListNode:
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.next = None
 6 
 7 class Solution:
 8     def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
 9         dummy = ListNode(0)
10         head = dummy
11         
12         carry = 0
13         
14         while l1 is not None or l2 is not None or carry == 1:
15             sum = carry
16             
17             if l1 is not None:
18                 sum += l1.val
19                 l1 = l1.next
20             if l2 is not None:
21                 sum += l2.val
22                 l2 = l2.next
23             
24             carry = math.floor(sum / 10)
25             curr = ListNode(sum % 10)
26             head.next = curr
27             head = head.next
28         
29         return dummy.next

标签：Medium,ListNode,sum,Two,next,Add,l2,l1,carry
来源： https://www.cnblogs.com/raymondwang/p/10890932.html

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