大型网站多少钱,企业咨询管理是干嘛的,分享公众号的网站,自己搭建服务器违法吗链表OJ 题目一#xff1a;移除链表元素题目二#xff1a;反转链表题目三#xff1a;链表的中间节点题目四#xff1a;链表中倒数第k个结点题目五#xff1a;合并两个有序链表题目六#xff1a;链表分割题目七#xff1a;链表的回文结构题目八#xff1a;相交链表题目九… 链表OJ 题目一移除链表元素题目二反转链表题目三链表的中间节点题目四链表中倒数第k个结点题目五合并两个有序链表题目六链表分割题目七链表的回文结构题目八相交链表题目九环形链表题目十环形链表II 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!!!
