合并链表相关的练习
目录
一、合并两个有序链表
二、两数相加
一、合并两个有序链表
将两个升序链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。
示例 1:
输入:l1 = [1,2,4], l2 = [1,3,4]
输出:[1,1,2,3,4,4]
示例 2:
输入:l1 = [], l2 = []
输出:[]
示例 3:
输入:l1 = [], l2 = [0]
输出:[0]
提示:
-
两个链表的节点数目范围是
[0, 50]
-
-100 <= Node.val <= 100
-
l1
和l2
均按 非递减顺序 排列
代码实现一(不设置哨兵位的头结点):
struct ListNode* mergeTwoLists(struct ListNode* list1, struct ListNode* list2)
{if (list1 == NULL)return list2;else if (list2 == NULL)return list1;
struct ListNode* newhead = NULL;struct ListNode* tail = NULL;struct ListNode* cur1 = list1;struct ListNode* cur2 = list2;while (cur1 && cur2){if (cur1->val < cur2->val){if (newhead == NULL){newhead = tail = cur1;}else{tail->next = cur1;tail = cur1;}cur1 = cur1->next;}else{if (newhead == NULL){newhead = tail = cur2;}else{tail->next = cur2;tail = cur2;}cur2 = cur2->next;}}if (cur1){tail->next = cur1;}if (cur2){tail->next = cur2;}return newhead;
}
代码实现二(设置哨兵位的头结点):
struct ListNode* mergeTwoLists(struct ListNode* list1, struct ListNode* list2)
{struct ListNode* guard = (struct ListNode*)malloc(sizeof(struct ListNode));guard->next = NULL;struct ListNode* tail = guard;struct ListNode* cur1 = list1;struct ListNode* cur2 = list2;while (cur1 && cur2){if (cur1->val < cur2->val){tail->next = cur1;tail = cur1;cur1 = cur1->next;}else{tail->next = cur2;tail = cur2;cur2 = cur2->next;}}if (cur1){tail->next = cur1;}if (cur2){tail->next = cur2;}struct ListNode* head = guard->next;free(guard);return head;
}
二、两数相加
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。
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]
提示:
-
每个链表中的节点数在范围
[1, 100]
内 -
0 <= Node.val <= 9
-
题目数据保证列表表示的数字不含前导零
代码实现:
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2)
{struct ListNode* guard = (struct ListNode*)malloc(sizeof(struct ListNode));guard->next = NULL;struct ListNode* tail = guard;// 两数相加int sum = 0;while (l1 || l2 || sum){if (l1 != NULL){sum += l1->val;l1 = l1->next;}if (l2 != NULL){sum += l2->val;l2 = l2->next;}// 生成一个新结点struct ListNode* newnode = (struct ListNode*)malloc(sizeof(struct ListNode));newnode->val = sum % 10; // newnode->val 设置为 sum 的个位newnode->next = NULL;// 尾插tail->next = newnode;tail = newnode; // 或者写成 tail = tail->next;// 更新 sumsum /= 10; // sum 更新为原来 sum 的十位}struct ListNode* head = guard->next;free(guard);return head;
}