02.01 移除重复节点

用unordered_set查重即可

02.02 返回倒数第k个节点

快慢指针

02.03 删除中间节点

一般情况下要用多个指针，但是可以直接消灭后一个，也算删除。
杀不掉我，我就变成你，然后再干掉你，就等于杀死了自己。

class Solution {
public:
    void deleteNode(ListNode* node) {
         node->val = node->next->val;
    node->next = node->next->next;
        
    }
};

02.04 分割链表

直接新开俩链表，暴力循环即可。
虚拟头节点，练习一下python

class Solution:
    def partition(self, head: ListNode, x: int) -> ListNode:
        dummy1, dummy2 = ListNode(), ListNode()
        cur1,cur2, cur = dummy1, dummy2, head
        while cur:
            if cur.val < x:
                cur1.next = cur
                cur1 = cur1.next
            else:
                cur2.next = cur
                cur2 = cur2.next
            cur = cur.next
        cur1.next, cur2.next = dummy2.next, None
        return dummy1.next

02.05 链表求和

暴力求解即可

02.06 回文链表

这个比较有意思
先用快慢指针将链表分成两段，再将后一段链表翻转。之后依次比较即可。

class Solution {
public:
    bool isPalindrome(ListNode* head) {
        if ( !head ) return true;
        ListNode* fast = head->next, *slow = head;
        while( fast && fast->next ){
            fast = fast->next->next;
            slow = slow->next;
        }
        // slow->next是后半段的链表的head   长度为5的话，slow在第三个，拆一下，32
        // 把后半段翻转
        ListNode* pre=NULL, *cur=slow->next, *sav;
        while(cur){
            sav = cur->next;
            cur->next = pre;
            pre = cur;
            cur = sav;
        }
        // 翻转结束
        // pre就是翻转后的head
        cur = pre;
        fast = head;
        while(cur&&fast&&cur->val==fast->val){
            cur = cur->next;
            fast = fast->next;
        }
        bool res = cur==NULL;
        // 还原链表
        /*
        cur = pre;
        pre = NULL;
        while(cur){
            sav = cur->next;
            cur->next = pre;
            pre = cur;
            cur = sav;
        }
        slow->next = pre;
        */
        return res;
    }
};

02.07 链表相交

这题真nb,凡是数学题都nb

在这里插入图片描述

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        ListNode *h1=headA, *h2=headB;
        while(h1!=h2)
        {
            h1=h1==nullptr? headB: h1->next;
            h2=h2==nullptr? headA: h2->next;
        }
        return h1;
    }
};

设交集链表长c,链表1除交集的长度为a，链表2除交集的长度为b，有
a + c + b = b + c + a
若无交集，则a + b = b + a

02.08 环路检测

常考
在这里插入图片描述

可以参考严格的数学证明
https://leetcode-cn.com/problems/linked-list-cycle-lcci/solution/kuai-man-zhi-zhen-zheng-ming-bi-jiao-yan-jin-by-ch/

class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        ListNode *slow = head, *fast = head;
        while (fast != nullptr) {
            slow = slow->next;
            if (fast->next == nullptr) {
                return nullptr;
            }
            fast = fast->next->next;
            if (fast == slow) {  //当快慢指针相遇后
                ListNode *ptr = head;
                while (ptr != slow) {
                    ptr = ptr->next;
                    slow = slow->next;
                }
                return ptr;
            }
        }
        return nullptr;
        
    }
};