LeetCodeOJ——2.Add Two Numbers

最新推荐文章于 2020-06-06 21:42:55 发布

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本文介绍了一种使用链表实现两个非负整数相加的方法。输入的两个链表以逆序方式存储数字，通过尾插法创建新的链表来表示相加后的结果。

Add Two Numbers

You are given two linked lists representing two non-negative numbers. 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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

解题思路：主要考链表和链表合并，尾插法（一开始用了头插法，尾插法链表的的建立，一开始建立了一个无效的头结点，返回的时候注意将这个无效的头结点删去）。

代码如下：

/**
 * 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) {
        ListNode* res = new ListNode(0);
        res->next = NULL;
        ListNode* q = res;
        int add = 0;
        while (l1!=NULL && l2!=NULL){
            ListNode *tmpNode=new ListNode((l1->val + l2->val + add) % 10);
            tmpNode->next = NULL;
            add = (l1->val + l2->val + add) / 10;
            q->next = tmpNode;
            q = q->next;
            l1=l1->next;
            l2=l2->next;
        }
        while (l1!=NULL){
            ListNode *tmpNode = new ListNode((l1->val + add) % 10);
            tmpNode->next = NULL;
            add = (l1->val + add) / 10;
            q->next = tmpNode;
            q = q->next;
            l1=l1->next;
        }
        while (l2!=NULL){
            ListNode *tmpNode = new ListNode((l2->val + add) % 10);
            tmpNode->next = NULL;
            add = (l2->val + add) / 10;
            q->next = tmpNode;
            q = q->next;
            l2=l2->next;
        }
        if(add!=0){
            ListNode *tmpNode = new ListNode(add);
            tmpNode->next = NULL;
            q->next = tmpNode;
            q = q->next;
        }

        ListNode *p = res;
        q = res->next;
        delete p;
        return q;
    }
};