2-Add Two Numbers

最新推荐文章于 2020-02-25 12:04:32 发布

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最新推荐文章于 2020-02-25 12:04:32 发布

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分类专栏： leetcode 文章标签： leetcode 2 addNumbers

本文链接：https://blog.youkuaiyun.com/lichen_yun/article/details/78290600

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难度：medium
类别：linked list

1.题目描述

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.

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

2.算法分析

注意对头结点首先进行处理，并且在最后的时候记得将head->next设置为NULL即可，详细思路可以见前面写过的博客
http://blog.youkuaiyun.com/lichen_yun/article/details/78224990
本题比前面写过的(⊙o⊙)容易，直接在前面那篇博客中将反转部分去掉即可。

3.代码实现

/**
 * 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) {
        stack<ListNode*> mystack;
        //  get reversed list l1
        ListNode* p = l1;
        //  sum correspond num, notice the carry
        int carry = 0;
        ListNode* head = NULL;
        if (l1 != NULL && l2 != NULL) {
            carry = (l1->val + l2->val) / 10;
            head = new ListNode((l1->val + l2->val)%10);
            l1 = l1->next;
            l2 = l2->next;
        }
        else if (l1 != NULL && l2 == NULL) {
            head = new ListNode(l1->val);
            l1 = l1->next;
        }
        else if (l1 == NULL && l2 != NULL) {
            head = new ListNode(l2->val);
            l2 = l2->next;
        }
        ListNode* result = head;
        int sum = 0;
        while (l1 != NULL || l2 != NULL) {
            if (l1 != NULL && l2 != NULL) {
                if (carry == 1) sum = l1->val + l2->val + 1;
                else sum = l1->val + l2->val;
                carry = sum / 10;
                head->next = new ListNode(sum % 10);
                l1 = l1->next;
                l2 = l2->next;
            }
            else if (l1 != NULL && l2 == NULL) {
                head->next = new ListNode((l1->val + carry)%10);
                carry = (l1->val + carry) / 10;
                l1 = l1->next;
            }
            else if (l1 == NULL && l2 != NULL) {
                head->next = new ListNode((l2->val + carry)%10);
                carry = (l2->val + carry) / 10;
                l2 = l2->next;
            }
            head = head->next;
        }
        //  if the final carry is one
        if (carry == 1) {
            head->next = new ListNode(1);
            head = head->next;
        }
        head->next = NULL;
        return result;
    }
};