2.Add Two Numbers

题目描述

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.

我的解法

实现代码

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class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode* dummyhead = new ListNode(0);
ListNode* p1 = dummyhead;
int x = 0, y = 0, carry = 0;
while (l1 || l2){
if (l1){
x = l1 -> val;
l1 = l1 -> next;
}
else
x = 0;
if (l2){
y = l2 -> val;
l2 = l2 -> next;
}
else
y = 0;
p1 -> next = new ListNode((x + y + carry) % 10);
carry = x + y >= 10 ? 1 : 0;
p1 = p1 -> next;
}
if (carry)
p1 -> next = new ListNode(1);
return dummyhead -> next;
}
};

可以写得更简洁一些:

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/**
 * 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 cur = 0;
        ListNode* dummyHead = new ListNode(0);
        ListNode* p = dummyHead;
        while (cur || l1 || l2){
            if (l1){
                cur += l1 -> val;
                l1 = l1 -> next;
            }
            if (l2){
                cur += l2 -> val;
                l2 = l2 -> next;
            }
            p -> next = new ListNode(cur % 10);
            p = p -> next;
            cur /= 10;
        }
        return dummyHead -> next;
    }
};