初阶数据结构：链表相关题目练习（补充）

1. 单链表相关练习题
- 1.1 移除链表元素
- 1.2 反转链表
- 1.3 链表的中间结点
- 1.4 链表的倒数第k个结点
- 1.5 合并两个有序链表
- 1.6 链表分割
- 1.7 链表的回文结构
- 1.8 相交链表
- 1.9 判断一个链表中是否有环
- 1.10 寻找环状链表相遇点
- 1.11 链表的深度拷贝

1. 单链表相关练习题

注：单链表结构上存在一定缺陷，所以链表相关的题目一般都针对与单链表。

1.1 移除链表元素

题目要求：
题目信息：

头节点head
移除值val

题目链接：
移除链表元素

方法（顺序处理法）：

思路：分析链表结构与结点所处的位置（是否为空链表，结点是否为头结点），分情况依次处理。

过程演示：
在这里插入图片描述

struct ListNode* removeElements4(struct ListNode* head, int val)
{
	struct ListNode* pre = NULL;
	struct ListNode* cur = head;

	while (cur)
	{
		if (cur->val == val)
		{
			//头删
			if (cur == head)
			{
				head = head->next;
				free(cur);
				cur = head;
			}
			else//中间删
			{
				pre->next = cur->next;
				free(cur);
				cur = pre->next;
			}
		}
		else
		{
			pre = cur;
			cur = cur->next;
		}
	}

	return head;
}

1.2 反转链表

题目要求：

题目链接：
反转链表

过程演示：

struct ListNode* reverseList(struct ListNode* head) 
{
    struct ListNode* pre = NULL;
    struct ListNode* mid = head;
    struct ListNode* cur = NULL;

    if(head)
    {
        cur = head->next;
    }

    while(mid)
    {
        mid->next = pre;
        pre = mid;
        mid = cur;
        if(cur)
        {
            cur = cur->next;
        }
    }

    return pre;
    
}

1.3 链表的中间结点

题目要求：

题目链接：
链表的中间结点

过程演示（快慢指针法）：

struct ListNode* middleNode(struct ListNode* head) 
{
    struct ListNode* fast = head;
    struct ListNode* slow = head;

    while(fast != NULL && fast->next != NULL)
    {
        fast = fast->next->next;
        slow = slow->next;
    }

    return slow;
}

1.4 链表的倒数第k个结点

题目要求：

题目链接：
倒数第k个结点

过程演示：

struct ListNode* FindKthToTail(struct ListNode* pListHead, int k ) 
{
    struct ListNode* cur = pListHead;
    struct ListNode* pre = pListHead;
    while(cur && k--)
    {
        cur = cur->next;
    }

    if(k > 0)
    {
        pre = NULL;
    }

    while(cur)
    {
        pre = pre->next;
        cur = cur->next;
    }

    return pre;
}

1.5 合并两个有序链表

题目要求：

题目链接：
合并两个链表

struct ListNode* mergeTwoLists(struct ListNode* list1, struct ListNode* list2) 
{
    struct ListNode* newhead = NULL;
    struct ListNode* cur = NULL;
    struct ListNode* newnode = NULL;

    if(list1 == NULL)
        return list2;

    if(list2 == NULL)
        return list1;

    while(list1 != NULL && list2 != NULL)
    {
        if(list1->val <= list2->val)
        {
            newnode = list1;
            list1 = list1->next;
        }
        else
        {
            newnode = list2;
            list2 = list2->next;
        }
        
        if(newhead == NULL)
        {
            newhead = newnode;
            newnode->next = NULL;
            cur = newhead;
        }
        else
        {
            //在遍历过程中，list1 或 list2 会等于 NULL
            cur->next = newnode;
            if(newnode != NULL)
            {
                newnode->next = NULL;
                cur = cur->next;
            }
        }
    }

    //有一个链表本就为空
    if(list1)
    {
        cur->next = list1;
    }

    if(list2)
    {
        cur->next = list2;
    }

    return newhead;
}

1.6 链表分割

题目要求：

题目链接：
合并两个链表

ListNode* BuyNewNode(int val)
    {
        ListNode* newnode = (ListNode*)malloc(sizeof(ListNode));
        newnode->val = val;
        newnode->next = nullptr;

        return newnode;
    }

    ListNode* partition(ListNode* pHead, int x) 
    {
        ListNode* newhead1 = nullptr;
        ListNode* end1 = nullptr;
        ListNode* newhead2 = nullptr;
        ListNode* end2 = nullptr;
        ListNode* cur = pHead;
        while(cur)
        {
            if(cur->val < x)
            {
                if(newhead1 == nullptr)
                {
                    newhead1 = BuyNewNode(cur->val);
                    end1 = newhead1;             
                }
                else 
                {
                    end1->next = BuyNewNode(cur->val);
                    end1 = end1->next;
                }
            }
            else 
            {
                if(newhead2 == nullptr)
                {
                    newhead2 = BuyNewNode(cur->val);
                    end2 = newhead2;             
                }
                else 
                {
                    end2->next = BuyNewNode(cur->val);
                    end2 = end2->next;
                }
            }
            cur = cur->next;
        }

        if(newhead1 == nullptr)
        {
            newhead1 = newhead2;
        }
        else 
        {
            end1->next = newhead2;
        }

        return newhead1;
    }

1.7 链表的回文结构

题目要求：

题目链接：
回文串

ListNode* reverse(ListNode* head)
    {
        ListNode* pre = nullptr;
        ListNode* mid = head;
        ListNode* cur = head->next;
        while (mid)
        {
            mid->next = pre;
            pre = mid;
            mid = cur;
            if (cur)
            {
                cur = cur->next;
            }
        }

        return pre;
    }

    bool chkPalindrome(ListNode* A)
    {
        //找相同，逆置
        //逐点比较
        //回文结构存在奇数个结点

        //获取中间结点
        ListNode* fast = A;
        ListNode* slow = A;
        while(fast && fast->next)
        {
            fast = fast->next->next;
            if(fast)
            {
                slow = slow->next;
            }
        }

        //表1
        ListNode* newhead2 = slow->next;
        //表2
        slow->next = nullptr;
        ListNode* newhead1 = reverse(A);
        if(fast)
        {
            newhead1 = newhead1->next;
        }


        while (newhead1 && newhead2 && newhead1->val == newhead2->val)
        {
            newhead1 = newhead1->next;
            newhead2 = newhead2->next;
        }

        if (newhead1 == nullptr && newhead2 == nullptr)
        {
            return true;
        }

        return false;
    }

1.8 相交链表

题目要求：
题目链接：
相交链表

void swap(struct ListNode** node1, struct ListNode** node2)
{
    struct ListNode* tmp = *node1;
    *node1 = *node2;
    *node2 = tmp;
}

struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB) 
{
    struct ListNode* short_list1 = headA;
    struct ListNode* short_list2 = headA;
    struct ListNode* long_list1 = headB;
    struct ListNode* long_list2 = headB;

    while(short_list1 && long_list1)
    {
        short_list1 = short_list1->next;
        long_list1 = long_list1->next;
    }

    if(short_list1)
    {
        swap(&short_list1, &long_list1);
        swap(&short_list2, &long_list2);
    }

    while(long_list1)
    {
        long_list1 = long_list1->next;
        long_list2 = long_list2->next;
    }

    //while((short_list2 && long_list2) || short_list2 != long_list2)
    while(short_list2 && long_list2 && short_list2 != long_list2)
    {
        long_list2 = long_list2->next;
        short_list2 = short_list2->next;
    }

    return short_list2;
}

1.9 判断一个链表中是否有环

题目要求：

题目链接：
判断是否有环

//逻辑步骤存疑
bool hasCycle(struct ListNode *head) 
{
    struct ListNode* fast = head;
    struct ListNode* slow = head;
	
	//err: while(fast && slow != fast)
    while(fast && fast->next)
    {
        fast = fast->next->next;
        slow = slow->next;
        if(slow == fast)
        {
            return true;
        }
    }

    return false;
}

1.10 寻找环状链表相遇点

题目要求：

题目链接：
寻找环状链表相遇点

思路依靠结论：一个结点从起始点，一个结点从相遇点（快慢指针相遇点），以同速行走（一次走一步），当他们再一次初次相遇时，此相遇结点即为入环点。

struct ListNode *detectCycle(struct ListNode *head) 
{
    //快慢指针确定首次相遇点
    //起始点与相遇点出发，同速移动，最后的相遇的点即为环的进入点
    struct ListNode* fast = head;
    struct ListNode* slow = head;

    //遍历寻找相遇点
    while(fast && fast->next)
    {
        fast = fast->next->next;
        slow = slow->next;

        if(fast == slow)
        {
            break;
        }
    }
	
	//判断是否为环状链表
    if(fast == NULL || fast->next == NULL)
    {
        return NULL;
    }

    slow = head;
    while(fast != slow)
    {
        fast = fast->next;
        slow = slow->next;
    }

    return fast;
}

1.11 链表的深度拷贝

题目要求：
>题目链接：
链表的深度拷贝

//思路：
//生成拷贝结点
//调整拷贝结点的指针
//还原原链表，链接拷贝结点
struct Node* BuyNewNode(int val)
{
    struct Node* newnode = (struct Node*)malloc(sizeof(struct Node));
    newnode->val = val;
    newnode->next = NULL;
    newnode->random = NULL;

    return newnode;
}

struct Node* copyRandomList(struct Node* head)
{
    struct Node* node1 = head;
	//判断是否为空链表
    if(head == NULL)
    {
        return head;
    }
    //创建新节点
    while (node1)
    {
        struct Node* newnode = BuyNewNode(node1->val);
        newnode->next = node1->next;
        node1->next = newnode;

        node1 = node1->next->next;
    }

    //调整新节点的随机指针
    struct Node* node2 = head->next;
    node1 = head;
    while (node2)
    {
        if (node1->random)
        {
            node2->random = node1->random->next;
        }

        node1 = node1->next->next;
        if (node2->next)
        {
            node2 = node2->next->next;
        }
        else
        {
            node2 = node2->next;
        }
    }

    //还原链表，链接新链表
    node1 = head;
    node2 = head->next;
    struct Node* newhead = head->next;
    while (node1)
    {
        node1->next = node1->next->next;
        if (node2->next)
        {
            node2->next = node2->next->next;
        }

        node1 = node1->next;
        node2 = node2->next;
    }

    return newhead;
}