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(牛客网)链表相加(二)

news 2025/9/13 18:03:57

嗯哼~

题目

描述

假设链表中每一个节点的值都在 0 - 9 之间，那么链表整体就可以代表一个整数。

给定两个这种链表，请生成代表两个整数相加值的结果链表。

数据范围：0 ≤ n,m ≤ 1000000，链表任意值 0 ≤ val ≤ 9
要求：空间复杂度 O(n)，时间复杂度 O(n)

示例

思路

单链表的翻转详细讲解：反转一个单链表（<---点击可看详解）

题解代码

struct ListNode* ReverseList(struct ListNode* pHead)
{// write code herestruct ListNode* cur = pHead;struct ListNode* pre = NULL;while (cur != NULL){struct ListNode* temp = cur->next;cur->next = pre;pre = cur;cur = temp;}return pre;
}struct ListNode* addInList(struct ListNode* head1, struct ListNode* head2)
{// write code herehead1 = ReverseList(head1);head2 = ReverseList(head2);int temp = 0;int add = 0;struct ListNode* newhead = NULL;while (head1 && head2){temp = (head1->val + head2->val + add);if (temp >= 10){add = 1;temp %= 10;}else{add = 0;}struct ListNode* cur = (struct ListNode*)malloc(sizeof(struct ListNode));cur->val = temp;cur->next = newhead;newhead = cur;head1 = head1->next;head2 = head2->next;}struct ListNode* empty = head1;struct ListNode* nonempty = head2;if (head1 != NULL){empty = head2;nonempty = head1;}while (nonempty){temp = (nonempty->val + add);if (temp >= 10){add = 1;temp %= 10;}else{add = 0;}struct ListNode* cur = (struct ListNode*)malloc(sizeof(struct ListNode));cur->val = temp;cur->next = newhead;newhead = cur;nonempty = nonempty->next;}if (add == 1){struct ListNode* cur = (struct ListNode*)malloc(sizeof(struct ListNode));cur->val = 1;cur->next = newhead;newhead = cur;}return newhead;}