Leetcode - 两数相加

题目

给你两个非空的链表，表示两个非负的整数。它们每位数字都是按照逆序的方式存储的，并且每个节点只能存储一位数字。

请你将两个数相加，并以相同形式返回一个表示和的链表。

你可以假设除了数字 0 之外，这两个数都不会以 0 开头。

样例

示例 1：

输入：l1 = [2,4,3], l2 = [5,6,4]

输出：[7,0,8]

解释：342 + 465 = 807.

示例 2：

输入：l1 = [0], l2 = [0]

输出：[0]

示例 3：

输入：l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]

输出：[8,9,9,9,0,0,0,1]

解决方案一

遵循加法法则，设置进位项，从低位到高位逐位相加。

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
    ListNode* head = new ListNode(), *p1 = l1, *p2 = l2, *p = head;
        int addition = 0;
        int sum = 0;
        while(p1 && p2){
            sum = p1->val + p2->val + addition; 
            addition = sum/10;
            p->next = new ListNode(sum % 10);
            p = p->next;
            p1 = p1->next;
            p2 = p2->next;
            }
        while(p1){
            sum = p1->val + addition; 
            addition = sum/10;
            p->next = new ListNode(sum % 10);
            p = p->next;
            p1 = p1->next;
        }
        while(p2){
            sum = p2->val + addition; 
            addition = sum/10;
            p->next = new ListNode(sum % 10);
            p = p->next;
            p2 = p2->next;
        }
        if(addition) p->next = new ListNode(addition);
        return head->next;
    }
};

1568/1568 cases passed (32 ms)

Your runtime beats 89.01 % of cpp submissions

Your memory usage beats 85.85 % of cpp submissions

解决方案二

利用递归将对应位数相加，最后添加进位项。

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        if (l1 == NULL and l2 == NULL) return NULL;
      else if (l1 == NULL) return l2; 
      else if (l2 == NULL) return l1; 

        int a = l1->val + l2->val;
        ListNode *p = new ListNode(a % 10);
        p->next = addTwoNumbers(l1->next,l2->next);
        if (a >= 10) p->next = addTwoNumbers(p->next, new ListNode(1));
        return p;
    }
};

1568/1568 cases passed (56 ms)

Your runtime beats 17.92 % of cpp submissions

Your memory usage beats 5.19 % of cpp submissions