代码随想录算法训练营day04|24.两两交换链表中的节点，19.删除链表的倒数第N个节点，面试题 02.07.链表相交，142.环形链表II

24.两两交换链表中的节点

题目链接：https://leetcode.cn/problems/swap-nodes-in-pairs/description/

我的代码：

/**
 * 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* dummy_head = new ListNode;
        dummy_head->next = head;
        ListNode* p = dummy_head;
        while (p->next != nullptr && p->next->next != nullptr) {
            ListNode* temp1 = p->next->next->next;
            ListNode* temp2 = p->next;
            p->next = p->next->next;
            p->next->next = temp2;
            p->next->next->next = temp1;
            p = p->next->next;
        }
        head = dummy_head->next;
        delete dummy_head;
        return head;
    }
};

还是定义虚拟头节点，同时注意循环条件，存储临时节点等细节问题。

19.删除链表的倒数第N个节点

题目链接：https://leetcode.cn/problems/remove-nth-node-from-end-of-list/description/

快慢指针法：

/**
 * 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* dummy_head = new ListNode;
        dummy_head->next = head;
        ListNode* fast = dummy_head;
        ListNode* slow = dummy_head;
        while (n--) {
            fast = fast->next;
        }
        while (fast->next) {
            fast = fast->next;
            slow = slow->next;
        }
        ListNode* p = slow->next;
        slow->next = p->next;
        delete p;
        head = dummy_head->next;
        delete dummy_head;
        return head;
    }
};

定义一个虚拟头节点，先让快指针走n步，之后两指针同时走直到fast的next节点为空，此时慢指针指向倒数第n个节点的前一个，即可执行删除。

面试题 02.07.链表相交

题目链接：https://leetcode.cn/problems/intersection-of-two-linked-lists-lcci/description/

我的代码：

/**
 * 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;
        int lengthB = 0;
        while (curA != nullptr) {
            lengthA++;
            curA = curA->next;
        }
        while (curB != nullptr) {
            lengthB++;
            curB = curB->next;
        }
        curA = headA;
        curB = headB;
        if (lengthA < lengthB) {
            swap(lengthA, lengthB);
            swap(curA, curB);
        }
        int gap = lengthA - lengthB;
        while (gap--) {
            curA = curA->next;
        }
        while (curA != nullptr) {
            if (curA == curB) {//注意是节点相等不是节点值相等，写成curA->val == curB->val会答案错误
                return curA;
            }
            curA = curA->next;
            curB = curB->next;
        }
        return NULL;//不存在相应节点时别忘了返回NULL
    }
};

求出两链表长度，将A链表置为较长链表，然后将curA移动到与curB平齐的位置，开始比较。

142.环形链表II

题目链接：https://leetcode.cn/problems/linked-list-cycle-ii/description/

我的代码：

/**
 * 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* fast = head;
        ListNode* slow = head;
        while (fast != nullptr && fast->next != nullptr) {
            fast = fast->next->next;
            slow = slow->next;
            if (fast == slow) {
                ListNode* index1 = head;
                ListNode* index2 = fast;
                while (index1 != index2) {
                    index1 = index1->next;
                    index2 = index2->next;
                }
                return index1;
            }
        }
        return NULL;
    }
};

快慢指针法判断有无环，fast走两步，slow走一步，若有环两指针必相遇。
对环入口位置的数学推导：