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leetcode-2-add two numbers

admin 11月 12, 2020 0

题目

给定两个链表，输出两个链表对应元素的和，返回结果链表

描述

举例说明
输入：(2 -> 4 -> 3) + (5 -> 6 -> 4)
输出：7 -> 0 -> 8

Java代码实现


 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class  {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode tmp = new ListNode(0);
        ListNode res = tmp;
        int x = 0;
        int sum;
        
        while(l1 != null || l2 != null || x != 0) {             //切记，不要忘了进位！！！！
            sum = x + (l1 != null ? l1.val : 0) + (l2 != null ? l2.val : 0);
            x = sum / 10;
            sum = sum % 10;
            tmp.next = new ListNode(sum);  //注意tmp.next是一个class，需要new进行新建
            l1 = l1 != null ? l1.next : null;
            l2 = l2 != null ? l2.next : null;
            tmp = tmp.next;
        }
        
        return res.next;
    }
}

Python代码实现

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class (object):
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        tmp = ListNode(0)
        result = tmp
        x = 0;
        while(x or l1 or l2):
            y1 = l1.val if l1 else 0
            y2 = l2.val if l2 else 0
            sum = (y1+y2+x)%10
            x = (y1+y2+x)/10
            result.next = ListNode(sum)
            result = result.next
            l1 = l1.next if l1 else None
            l2 = l2.next if l2 else None
        return tmp.next

C++代码实现

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode tmp(0);
        struct ListNode *result = &tmp;
        int x = 0;
        int sum;
        while(l1 || l2 || x){
            sum = x+(l1?l1->val:0)+(l2?l2->val:0);
            x = sum/10;
            sum = sum%10;
            result->next = new ListNode(sum);
            result = result->next;
            l1 = l1 ? l1->next: NULL;
            l2 = l2 ? l2->next: NULL;
        }
        return tmp.next;
    }
};