Day4：22两两交换链表中的结点 19删除链表的倒数第n个结点 面试02.07链表相交 142.环形链表

22l两两交换链表中的结点

while(cur->next != nullptr && cur->next->next != nullptr) 继续进行链表交换的条件是cur指针下一个结点不为空且下下个结点也不为空。如果只剩下一个结点（即链表中是奇数个结点）也没必要交换了。
时间复杂度O(n)
空间复杂度O（1）

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        ListNode* dummyhead = new ListNode(0);
        dummyhead->next = head;
        ListNode* cur = dummyhead;
        while(cur->next != nullptr && cur->next->next != nullptr){
            ListNode* temp=cur->next;
            ListNode* temp1=cur->next->next->next;
            cout<<"jiaohuan"<<temp->val<<endl;
            //cout<<"jiaohuan"<val<
            cur->next=cur->next->next;
            cur->next->next=temp;
            temp->next=temp1;
            cout<<"jiaohuan"<<cur->val<<endl;
            cout<<"jiaohuan"<<cur->next->val<<endl;
            cout<<"jiaohuan"<<cur->next->next->val<<endl;
            cur=cur->next->next;
            cout<<"jiaohuan"<<cur->val<<endl;
        }
        return dummyhead->next;
    }
};

删除列表倒数第N个结点

当下反应的思路是先遍历链表，算链表的长度，再算倒数第N个结点是正数第几个，但这样时间复杂度是o(2n)即遍历了两边。
拓展版用双指针只对链表遍历了一遍，时间复杂度O(N)，让快指针先走N步，或者N-1步（根据while判断条件的不同），然后slow指针再一起向前走，这样，随着快指针走到终点，保证慢指针走到删除结点的前一个，即可进行操作了，记得借助temp指针先保存要删除的结点，再进行cur->next=cur->next->next;
空间复杂度都是O（1）

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode* dummyhead = new ListNode(0);
        dummyhead->next = head;
        ListNode* slow = dummyhead;
        ListNode* fast = dummyhead;
        while(n-- && fast->next != nullptr){
            fast = fast->next;
        }
        while(fast->next != nullptr){
            slow=slow->next;
            fast=fast->next;
        }
        ListNode* temp=slow->next;
        slow->next=slow->next->next;
        delete temp;

        return dummyhead->next;
    }
};

链表相交

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        ListNode* curA = headA;
        ListNode* curB = headB;
        int LengthA=0,LengthB=0;

        while(curA->next != NULL){
            curA=curA->next;
            LengthA++;
        }
        while(curB->next != NULL){
            curB=curB->next;
            LengthB++;
        }
        curA=headA;curB=headB;
        if(LengthA<LengthB){
            swap(LengthA,LengthB);
            swap(curA,curB);
        }
        int gap=LengthA-LengthB;
        while(gap--){
            curA=curA->next;
        }
        while(curA != NULL && curB != NULL && curA!=curB){
            curA=curA->next;
            curB=curB->next;
        }
        return curA;
    }
};

142环形链表

关于什么时候return null还是需要多思考一下详细见代码注释

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        ListNode* slow = head;
        ListNode* fast = head;

        while(fast != NULL && fast->next != NULL ){
            slow = slow->next;
            fast = fast->next->next; 
            if(slow == fast){
                ListNode* index=head;
                ListNode* index1=fast;

                while(index != index1){
                    index=index->next;
                    index1=index1->next;
                }
                return index;
            }
            //else 如果在这里就return null的话相当每一次循环的快慢指针不相等都会return null
            	//return NULL;
        }
        return NULL;
    
        }
};