leetcode 4 ADD TWO NUMBERS

最新推荐文章于 2018-09-18 00:34:56 发布

mlweixiao

最新推荐文章于 2018-09-18 00:34:56 发布

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分类专栏： leetcode LeetCode专栏文章标签： leetcode c++

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本文介绍了一种算法问题的解决方案：两个非负整数由链表逆序存储，如何将这两个数相加并返回一个新的链表。提供了两种C++实现方式，一种使用头结点，另一种则不使用。

Youare given two linked lists representing two non-negative numbers. The digitsare stored in reverse order and each of their nodes contain a single digit. Addthe two numbers and return it as a linked list.

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

C++ 有头结点版

/**
 * 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 rootNode(0);
        ListNode *pCurNode = &rootNode;
        int v1,v2,sum;
        int forward = 0;
        while(l1||l2)
        {
            v1 = (l1 ? l1->val : 0);
            v2 = (l2 ? l2->val : 0);
            sum = v1 + v2 + forward;
            forward = sum / 10;
            sum = sum % 10;
            ListNode *pNode = new ListNode(sum);
            pCurNode->next = pNode;
            pCurNode = pNode;
            if(l1) l1 = l1->next;
            if(l2) l2 = l2 ->next;
        }
        if(forward > 0)
        {
            ListNode *pNode = new ListNode(forward);
            pCurNode->next = pNode;
        }
        return rootNode.next;
    }
};

C++ 无头结点版

/**
 * 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 v1 = (l1 ? l1->val : 0);
        int v2 = (l2 ? l2->val : 0);
        int sum = v1 + v2;
		ListNode *pCurNode =new ListNode(sum%10);
		ListNode *p=pCurNode;
		int forward = sum/10;
		
		if(l1) l1 = l1->next;
        if(l2) l2 = l2 ->next;
        
        while(l1||l2)
        {
            v1 = (l1 ? l1->val : 0);
            v2 = (l2 ? l2->val : 0);
            sum = v1 + v2 + forward;
            forward = sum / 10;
            sum = sum % 10;
            ListNode *pNode = new ListNode(sum);
            pCurNode->next = pNode;
            pCurNode = pNode;
            
            if(l1) l1 = l1->next;
            if(l2) l2 = l2 ->next;
        }
        if(forward > 0)
        {
            ListNode *pNode = new ListNode(forward);
            pCurNode->next = pNode;
        }
        return p;
    }
};