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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：

1. 头节点head
2. 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：

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 NULLcur->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1ListNode* newhead2 = slow->next;//表2slow->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;
}