2. Add Two Numbers

最新推荐文章于 2021-12-02 14:05:46 发布

原创最新推荐文章于 2021-12-02 14:05:46 发布 · 111 阅读

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本文介绍了一种算法，用于解决两个链表表示的大数相加的问题。该算法使用C++和Python实现，通过遍历两个链表，逐位相加并处理进位，最终返回结果链表。

遍历链表：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) 
    {
        int flag=0,temp;
        ListNode *res=l1;
        ListNode *last;
        while(l1!=NULL&&l2!=NULL)
        {
            temp=l1->val + l2->val + flag;
            l1->val=temp%10;
            flag=temp/10;
            last=l1;
            l1=l1->next;
            l2=l2->next;
        }
        while(l1!=NULL)
        {
            temp=l1->val + flag;
            l1->val=temp%10;
            flag=temp/10;
            last=l1;
            l1=l1->next;
        }
        
        last->next=l2;
         while(l2!=NULL)
        {
            temp=l2->val + flag;
            l2->val=temp%10;
            flag=temp/10;
             last=l2;
            l2=l2->next;
        }
        
        if(flag>0)
        {
            ListNode *node=(ListNode*)malloc(sizeof(ListNode));
            node->val=flag;
            node->next=NULL;
            last->next = node;
        }
        return res;
    }
};

python 版：

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

class Solution:
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        l3 = ListNode(None)
        last = ListNode(None)
        l3 = l1
        jin = 0
        while l1 and l2:
            l1.val += (l2.val + jin)
            jin = l1.val // 10
            l1.val = l1.val % 10
            
            last = l1
            l1 = l1.next
            l2 = l2.next

        if l1:
            while jin and l1:
                l1.val += jin
                jin = l1.val // 10
                l1.val = l1.val % 10
                last = l1
                l1 = l1.next

        if l2:
            last.next = l2
            while jin and l2:
                l2.val += jin
                jin = l2.val // 10
                l2.val = l2.val % 10
                last = l2
                l2 = l2.next

        if jin:
            newNode = ListNode(jin)
            last.next = newNode
        return l3