LeetCode 002 Add Two Numbers

飞行荷兰人

于 2019-04-02 20:55:27 发布

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本文解决LeetCode 2.两数相加问题，实现链表节点的遍历与相加操作，考虑了不同长度链表的处理及进位情况。提供两种解决方案，一种为详细步骤实现，另一种为简洁版代码。

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题目：LeetCode 2. Add Two Numbers

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.

这是一个简单的链表遍历，注意链表的长短，避免空指针，以及注意进位。

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) {
 		int carry = 0;
        int num=0;
		ListNode* head;
		ListNode* add1=l1;
		ListNode* add2=l2;
		ListNode * current = head = new ListNode(0);

		current->val = (add1->val + add2->val) % 10;
		carry = (add1->val + add2->val) / 10;
		add1 = add1->next;
		add2 = add2->next;

		while (add1 != NULL || add2 != NULL)
		{
			if (add1 != NULL && add2 != NULL)
			{
				num = (add1->val + add2->val + carry)%10;
				carry = (add1->val + add2->val + carry) / 10;
				add1 = add1->next;
				add2 = add2->next;
			}
			else if (add1 != NULL)
			{
				num = (add1->val+ carry) % 10;
				carry = (add1->val + carry) / 10;
				add1 = add1->next;
			}
			else if (add2 != NULL)
			{
				num = (add2->val + carry) % 10;
				carry = ( add2->val + carry) / 10;
				add2 = add2->next;
			}
            current->next = new ListNode(num);
			current = current->next;
		}
        if(carry>0)
			current->next = new ListNode(carry);
		return head;
    }
};

这题没什么好说的，有兴趣的可以研究下讨论区里大佬的11行代码：

[c++] Sharing my 11-line c++ solution, can someone make it even more concise?

ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
    ListNode preHead(0), *p = &preHead;
    int extra = 0;
    while (l1 || l2 || extra) {
        int sum = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + extra;
        extra = sum / 10;
        p->next = new ListNode(sum % 10);
        p = p->next;
        l1 = l1 ? l1->next : l1;
        l2 = l2 ? l2->next : l2;
    }
    return preHead.next;
}