1.单链表反转

public ListNode reverse(ListNode head) {
    if (head == null || head.next == null) {
        return head;
    }
    ListNode preNode = null;
    ListNode nextNode = null;
    while (head != null){
        nextNode = head.next;
        head.next = preNode;
        preNode = head;
        head = nextNode;
    }
    return preNode;
}

2.链表中环的检测

// 快慢指针解法
public boolean detectLoop(ListNode head) {
        ListNode pSlow = head, pFast = head;
        boolean detectFlag = false;
        if (head.next == null) {
            return detectFlag;
        }
        while (true) {
            pSlow = pSlow.next;
            pFast = pFast.next.next;
            if (pFast == null) {
                break;
            }
            if (pSlow == pFast) {
                detectFlag = true;
                break;
            }
        }
        return detectFlag;
    }

3.两个有序的链表合并

// 遍历解法
// 同时不断遍历两个链表，取出小的追加到新的头节点后，直至两者其中一个为空
// 再将另一者追加的新链表最后
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
     ListNode dummy = new ListNode(-1);
     ListNode curNode = dummy;
     while (l1 != null && l2 != null) {
         if (l1.val <= l2.val) {
             curNode.next = l1;
             l1 = l1.next;
         } else {
             curNode.next = l2;
             l2 = l2.next;
         }
         curNode = curNode.next;
     }
     curNode.next = (l1 != null) ? l1 : l2;
     return dummy.next;
 }

4.删除链表倒数第n个结点

public ListNode removeNthFromEnd(ListNode head, int n) {
    ListNode preNode = head;
    ListNode curNode = head;
    // 前指针先走nbu
    for (int i = 0; i < n; i++) {
        curNode = curNode.next;
    }
    if (curNode == null) {
        return preNode.next;
    }
    // 前后指针同时走，走到尾部，preNode.next就是需要删除的节点
    while (curNode.next != null) {
        preNode = preNode.next;
        curNode = curNode.next;
    }
    // 删除节点
    preNode.next = preNode.next.next;
    return head;
}

5.求链表的中间结点

// 使用快慢指针，一个指针每次只遍历一个节点，另一个速度为2倍，当快指针指向表尾时，慢指针指向中间节点
public ListNode middleNode(ListNode head) {
        ListNode slow = head, fast = head;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }