题目一：移除链表元素

OJ

方案一：

题目解析：

代码演示：
struct ListNode
{int val;struct ListNode* next;
};
struct ListNode* removeElements(struct ListNode* head, int val)
{struct ListNode* cur = head, * prev = NULL;while (cur){if (cur->val == val){if (cur == head){head = cur->next;free(cur);cur = head;}else{prev->next = cur->next;free(cur);cur = prev->next;}}else{prev = cur;cur = cur->next;}}return head;
}

方案二：

题目解析：把原链表遍历一遍，插入新链表

struct ListNode* removeElements(struct ListNode* head, int val)
{struct ListNode* newnode = NULL, * tail = NULL;struct ListNode* cur = head;while (cur){if (cur->val == val){struct ListNode* del = cur;cur = cur->next;free(del);}else{if (tail == NULL){newnode = tail = cur;//tail = tail->next;}else{tail->next = cur;tail = tail->next;}cur = cur->next;}}if (tail)tail->next = NULL;return newnode;
}

题目二：反转链表

OJ

题目解析：

代码演示：
struct ListNode
{int val;struct ListNode* next;
};
struct ListNode* reverseList(struct ListNode* head) 
{struct ListNode* newnode = NULL, * cur = head;while (cur){struct ListNode* next = cur->next;//尾插cur->next = newnode;newnode = cur;cur = next;}return newnode;
}

题目三：链表的中间节点

OJ

题目解析：

代码演示：
struct ListNode 
{int val;struct ListNode* next;
};
struct ListNode* middleNode(struct ListNode* head)
{struct ListNode* fast = head, * slow = head;while (fast && fast->next){slow = slow->next;fast = fast->next->next;}return slow;
}

题目四：链表中倒数第k个结点

OJ

题目解析：

代码演示：
struct ListNode
{int val;struct ListNode* next;
};
struct ListNode* FindKthToTail(struct ListNode* pListHead, int k) 
{struct ListNode* fast = pListHead, * slow = pListHead;while (k--){if (fast == NULL)return NULL;fast = fast->next;}while (fast){slow = slow->next;fast = fast->next;}return slow;
}

题目五：合并两个有序链表

OJ

题目解析：

代码演示：
struct ListNode 
{int val;struct ListNode* next;
};
struct ListNode* mergeTwoLists(struct ListNode* list1, struct ListNode* list2) 
{if (list1 == NULL)return list2;if (list2 = NULL)return list1;struct ListNode* tail = NULL, * head = NULL;head = tail = (struct ListNode*)malloc(sizeof(struct ListNode));while (list1 && list2){if (list1->val < list2->val){tail->next = list1;tail = tail->next;list1 = list1->next;}else{tail->next = list2;tail = tail->next;list2 = list2->next;}}if (list1)tail->next = list1;if (list2)tail->next = list2;struct ListNode* del = head;head = head->next;free(del);return head;
}

题目六：链表分割

OJ

题目解析：

代码演示：
struct ListNode 
{int val;struct ListNode* next;ListNode(int x) : val(x), next(NULL) {}
}; 
class Partition
{
public:ListNode* partition(ListNode* pHead, int x) {struct ListNode* lhead, * ltail, * gtail, * ghead;ghead = gtail = (struct ListNode*)malloc(sizeof(struct ListNode));lhead = ltail = (struct ListNode*)malloc(sizeof(struct ListNode));struct ListNode* cur = pHead;while (cur){if (cur->val < x){ltail->next = cur;ltail = ltail->next;}else{gtail->next = cur;gtail = gtail->next;}cur = cur->next;}ltail->next = ghead->next;gtail->next = NULL;struct ListNode* head = lhead->next;free(lhead);free(ghead);return head;}
};

题目七：链表的回文结构

OJ

题目解析：

代码演示：
struct ListNode 
{int val;struct ListNode* next;ListNode(int x) : val(x), next(NULL) {}
}; 
class PalindromeList
{
public:struct ListNode* middleNode(struct ListNode* head){struct ListNode* fast = head, * slow = head;while (fast && fast->next){slow = slow->next;fast = fast->next->next;}return slow;}struct ListNode* reverseList(struct ListNode* head){struct ListNode* newnode = NULL, * cur = head;while (cur){struct ListNode* next = cur->next;cur->next = newnode;newnode = cur;cur = next;}return newnode;}bool chkPalindrome(ListNode* A){struct ListNode* mid = middleNode(A);struct ListNode* rmid = reverseList(mid);while (A && rmid){if (A->val != rmid->val)return false;A = A->next;rmid = rmid->next;}return true;}
};

题目八：相交链表

OJ

题目解析：
定义快慢指针，使快指针先走与慢指针同步。然后同时走看是否相交

代码演示：
struct ListNode
{int val;struct ListNode* next;
};
struct ListNode* getIntersectionNode(struct ListNode* headA, struct ListNode* headB) 
{struct ListNode* curA = headA, * curB = headB;int lenA = 1, lenB = 1;while (curA){curA = curA->next;lenA++;}while (curB){curB = curB->next;lenB++;}if (curA != curB)return false;int gab = abs(lenA - lenB);struct ListNode* longest = headA, * shortest = headB;if (lenA < lenB){longest = headB;shortest = headA;}while (gab--){longest = longest->next;}while (shortest != longest){shortest = shortest->next;longest = longest->next;}return longest;
}

题目九：环形链表

OJ

题目解析：

代码演示：
struct ListNode {int val;struct ListNode* next;
};
bool hasCycle(struct ListNode* head)
{struct ListNode* fast = head, * slow = head;while (fast && fast->next){slow = slow->next;fast = fast->next->next;if (slow == fast)return true;}return false;
}

题目十：环形链表II

OJ

题目解析：

代码演示：
struct ListNode
{int val;struct ListNode* next;
};
struct ListNode* detectCycle(struct ListNode* head)
{struct ListNode* fast = head, * slow = head;while (fast && fast->next){slow = slow->next;fast = fast->next->next;if (slow == fast){struct ListNode* meet = slow;while(meet != head){head = head->next;meet = meet->next;}return meet;}}return NULL;
}

💘不知不觉，【数据结构初阶】链表OJ以告一段落。通读全文的你肯定收获满满，让我们继续为数据结构学习共同奋进!!!