单链表经典题目

1. 题目链接：合并两个链表力扣https://leetcode.cn/problems/merge-two-sorted-lists/

class Solution {
public:
    ListNode* mergeTwoLists(ListNode* list1, ListNode* list2) {
        ListNode* head = new ListNode(-1);//哨兵
        ListNode*tail = head;
        tail->next=NULL;
        while(list1&&list2)
        {
            if(list1->valval)
            {
                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;
        }
        return head->next;
    }
};

带哨兵：较为简单

不带哨兵 ：较为复杂，就需要考虑头结点为空的情况

2.找倒数第K个节点：使用 快慢指针

链表中倒数第k个结点_牛客题霸_牛客网【牛客题霸】收集各企业高频校招笔面试题目，配有官方题解，在线进行百度阿里腾讯网易等互联网名企笔试面试模拟考试练习,和牛人一起讨论经典试题,全面提升你的技术能力https://www.nowcoder.com/practice/529d3ae5a407492994ad2a246518148a?tpId=13&&tqId=11167&rp=2&ru=/activity/oj&qru=/ta/coding-interviews/question-ranking

class Solution {
public:
    ListNode* FindKthToTail(ListNode* pListHead, unsigned int k) {
        struct ListNode* slow ,*fast;
        fast = slow = pListHead;
        while(k--)
        {
            if(fast==NULL)
                return  NULL;
            fast = fast->next;
        }
        while(fast!=NULL)
        {
            slow = slow->next;
            fast = fast ->next;
        }
        //快慢指针 ： 先走k步，然后fast slow 同时走，当fast==NULL
        return slow;
    }
};

3.分割：链表分割_牛客题霸_牛客网【牛客题霸】收集各企业高频校招笔面试题目，配有官方题解，在线进行百度阿里腾讯网易等互联网名企笔试面试模拟考试练习,和牛人一起讨论经典试题,全面提升你的技术能力https://www.nowcoder.com/practice/0e27e0b064de4eacac178676ef9c9d70?tpId=8&&tqId=11004&rp=2&ru=/activity/oj&qru=/ta/cracking-the-coding-interview/question-ranking

思路：

定义两个链表，四个节点指针 ：头 尾 

class Partition {
public:
    ListNode* partition(ListNode* pHead, int x) {
        // write code here
        ListNode* smallHead = new ListNode(-1);
        ListNode* bigHead = new ListNode(-1);
        ListNode* pstail =smallHead,*pbtail = bigHead,*cur = pHead;
        while(cur)
        {
            if(cur->val < x)
            {
                pstail->next = cur;
                pstail = pstail->next;//移到到新节点 
            }else{
                pbtail->next = cur;
                pbtail = pbtail->next;
            }
            cur = cur->next;
        }
        pbtail->next=NULL;
        pstail->next = bigHead->next;
        return smallHead->next;
    }
};