代码随想录算法训练营day4|24. 两两交换链表中的结点、19. 删除链表的倒数第 N 个结点、面试题02.07.链表相交、142.环形链表II

1. 两两交换链表中的结点

题目链接：24. 两两交换链表中的结点
题目解析：https://programmercarl.com/0024.%E4%B8%A4%E4%B8%A4%E4%BA%A4%E6%8D%A2%E9%93%BE%E8%A1%A8%E4%B8%AD%E7%9A%84%E8%8A%82%E7%82%B9.html

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode swapPairs(ListNode head) {
        ListNode vir = new ListNode(); //虚拟头节点
        vir.next = head;  
        ListNode temp = null; //临时节点
        ListNode cur = vir;  //当前节点

        while (cur.next != null && cur.next.next != null) { //当当前节点后面还有两个结点非空则可以进行交换
            temp = cur.next.next;
            cur.next.next = temp.next;
            temp.next = cur.next;
            cur.next = temp;
            cur = cur.next.next;
        }
        return vir.next;

    }
}

用虚拟头结点，这样会方便很多。

2. 删除链表的倒数第 N 个结点

题目链接：19.删除链表的倒数第 N 个结点
题目解析：https://programmercarl.com/0019.%E5%88%A0%E9%99%A4%E9%93%BE%E8%A1%A8%E7%9A%84%E5%80%92%E6%95%B0%E7%AC%ACN%E4%B8%AA%E8%8A%82%E7%82%B9.html

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode vir = new ListNode();
        vir.next = head;
        ListNode start = vir;
        ListNode end = head;
        int count = 1;
        while (count < n) {
            end = end.next;
            count++;
        }
        while (end.next != null) {
            start = start.next;
            end = end.next;
        }
        start.next = start.next.next;
        return vir.next;

    }
}

3. 链表相交

题目链接：面试题02.07.链表相交
题目解析：https://programmercarl.com/%E9%9D%A2%E8%AF%95%E9%A2%9802.07.%E9%93%BE%E8%A1%A8%E7%9B%B8%E4%BA%A4.html

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        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) {
            //1. swap (lenA, lenB);
            int tmpLen = lenA;
            lenA = lenB;
            lenB = tmpLen;
            //2. swap (curA, curB);
            ListNode tmpNode = curA;
            curA = curB;
            curB = tmpNode;
        }
        // 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-- > 0) {
            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;
    }

}

4. 环形链表II

题目链接：142.环形链表II
题目解析：https://programmercarl.com/0142.%E7%8E%AF%E5%BD%A2%E9%93%BE%E8%A1%A8II.html

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public 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 index1 = fast;
                ListNode index2 = head;
                while (index1 != index2) {
                    index1 = index1.next;
                    index2 = index2.next;
                }
                return index1;
            }
        }
        return null;
    }
}