Leetcode | Add Two Numbers

You are given two linked lists representing two non-negative numbers. 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.

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

注意取指针时要判断是否为空。

一般尾插入的话，用一个dummy结点。这样就可以不用特殊处理头指针了。

/**
 * 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 ret(0);
        ListNode *p = &ret;
        int carry = 0, v1, v2, s;
        
        while (l1 != NULL || l2 != NULL) {
            v1 = l1 ? l1->val : 0;
            v2 = l2 ? l2->val : 0;
            s = v1 + v2 + carry;
            carry = s / 10;
            p->next = new ListNode(s % 10);
            p = p->next;
            if (l1) l1 = l1->next;
            if (l2) l2 = l2->next;
        }
        if (carry > 0) {
            p->next = new ListNode(carry);
        }
        return ret.next;
    }
};

 代码现在是越写越简洁了，好事。逻辑清晰了许多。

/**
 * 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, v = 0;
        ListNode h(0), *p = &h;
        while (l1 || l2) {
            v = carry;
            if (l1) {
                v += l1->val;
                l1 = l1->next;
            } 
            if (l2) {
                v += l2->val;
                l2 = l2->next;
            }
            p->next = new ListNode(v % 10);
            p = p->next;
            carry = v / 10;
        }
        if (carry > 0) {
            p->next = new ListNode(carry);
        }
        return h.next;
    }
};