代码随想录算法训练营第四天 | 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、 链表总结

24. 两两交换链表中的节点

文章：代码随想录 (programmercarl.com)

视频：https://www.bilibili.com/video/BV1YT411g7br

class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        ListNode* dummyHead = new ListNode(0);
        dummyHead->next = head;
        ListNode* temp1;
        ListNode* temp2;
        ListNode* cur = dummyHead;

        while (cur->next != NULL && cur->next->next != NULL)
        {
            temp1 = cur->next; //记录结点1
            temp2 = cur->next->next->next;//记录结点3
            
            cur->next = cur->next->next;//dummyHead->2
            cur->next->next = temp1;//2->1
            cur->next->next->next = temp2;//1->3

            cur = cur->next->next; //cur移至下一组结点的前一位
        }
        return dummyHead->next; //返回头结点
    }
};

19. 删除链表的倒数第 N 个结点

文章：代码随想录 (programmercarl.com)

视频：https://www.bilibili.com/video/BV1vW4y1U7Gf

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode* dummyHead = new ListNode(0);
        dummyHead->next = head;
        ListNode* fastNode = dummyHead;
        ListNode* slowNode = dummyHead;
        ListNode* temp;
        
        n++; //为了让slow指针停在待删除节点前一位
        while (n-- != 0 && fastNode != NULL) {
            //快指针 - 慢指针 = n+1
            //快指针为NULL时，慢指针在待删除节点前一位
            fastNode = fastNode->next; 
        }
        
        while (fastNode != NULL) {
            fastNode = fastNode->next;
            slowNode = slowNode->next;
        }

        temp = slowNode->next;
        slowNode->next = slowNode->next->next;
        delete temp;

        return dummyHead->next;
    }
};

160. 相交链表

文章：代码随想录 (programmercarl.com)

ListNode* curA = headA;
        ListNode* curB = headB;
        int lenA = 0, lenB = 0;
        while (curA != NULL) { // 求链表A的长度
            lenA++;
            curA = curA->next;
        }
        while (curB != NULL) { // 求链表B的长度
            lenB++;
            curB = curB->next;
        }
        curA = headA;
        curB = headB;
        // 让curA为最长链表的头，lenA为其长度
        if (lenB > lenA) {
            swap (lenA, lenB);
            swap (curA, curB);
        }
        // 求长度差
        int gap = lenA - lenB;
        // 让curA和curB在同一起点上（末尾位置对齐）
        while (gap--) {
            curA = curA->next;
        }
        // 遍历curA 和 curB，遇到相同则直接返回
        while (curA != NULL) {
            if (curA == curB) {
                return curA;
            }
            curA = curA->next;
            curB = curB->next;
        }
        return NULL;

142. 环形链表 II

文章：代码随想录 (programmercarl.com)

视频：https://www.bilibili.com/video/BV1if4y1d7ob

class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        ListNode* slow = head;
        ListNode* fast = head;

        while (fast != NULL && fast->next != NULL)
        {
            fast = fast->next->next;
            slow = slow->next;

            if (fast == slow)
            {
                ListNode* index1 = fast;
                ListNode* index2 = head;
                while (index1 != index2)
                {
                    index1 = index1->next;
                    index2 = index2->next;
                }
                return index2; //返回环的入口
            }
        }
        return NULL;
    }
};

//数学推导
slow = x + y; //慢指针每次前进一个单位
fast = x + y + n(y + z); //快指针每次前进一个单位

2(x + y) = x + y + n(y + z);
x + y = n(y + z);
x = (n - 1)(y + z) + z;