[LeetCode] Add Two Numbers

最新推荐文章于 2022-07-02 19:53:18 发布

dongyu1703

最新推荐文章于 2022-07-02 19:53:18 发布

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本文介绍了一种算法，用于处理两个非空链表表示的非负整数相加的问题。链表中的数字以逆序存储，每个节点包含一个单独的数字。通过示例解释了如何将两个数字相加并返回结果链表。

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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.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

// 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 addnum = 0;
		int len1 = 0;
		int len2 = 0;
		ListNode* temp = l1;
		ListNode* result = new ListNode(0);
		while (temp != NULL) {
			len1 += 1;
			temp = temp->next;
		}
		temp = l2;
		while (temp != NULL) {
			len2 += 1;
			temp = temp->next;
		}
		int len = len1 < len2 ? len2 : len1;
		temp = result;
		ListNode* a = l1;
		ListNode* b = l2;
		int sum = 0;
		for (int i = 0; i < len; i++) {
			if (a == NULL)
				sum = b->val;
			else if (b == NULL)
				sum = a->val;
			else 
				sum = a->val + b->val;
			sum = sum + addnum;
			if (sum > 9) {
				sum = sum % 10;
				addnum = 1;
			} else
				addnum = 0;
			temp->val = sum;
			if (i != len - 1)
				temp->next = new ListNode(0);
			temp = temp->next;
			if (a != NULL)
				a = a->next;
			if (b != NULL)
				b = b->next;
		}
		if (addnum == 1) {
			temp = result;
			while (temp->next != NULL)
				temp = temp->next;
			temp->next = new ListNode(addnum);
		}	
		return result;
	}
};