1.快慢指针

  先看一道简单的题目：返回中间结点

这道题有一个最朴素的做法就是先遍历一边链表，设置计数器求出链表长度，再重新走1/2的链表长度，即可返回中间节点

        // 第二种解法  
        //这种解法需要遍历两次链表
        ListNode cur1 = head;
        int cnt = 0;

        while(cur1 != null) {
            cnt++;
            cur1 = cur1.next;
        }

        ListNode cur2 = head;
        cnt = cnt/2;
        while(cnt != 0) {
            cur2 = cur2.next;
            cnt--;
        }

        return cur2;

  但是这种方式有个明显的缺陷，就是你实际上是遍历了两遍链表，那有没有只遍历一次链表就能获得中间结点的方法呢？答案是有的，利用快慢指针

// 第一种解法  只需遍历一次链表
        ListNode slow = head,fast = head;

        while(fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }

        return slow;

快慢指针的核心思想其实是一种数学问题，即在相同时间内，路程之比就是速度之比 

「力扣」第 19 题： 倒数第 k 个结点 

 https://leetcode.cn/problems/remove-nth-node-from-end-of-list/

class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        // 使用虚拟头节点 -- 能更好的处理删除头节点的问题
        ListNode dummyhead = new ListNode(0);
        dummyhead.next = head;

        ListNode slow = dummyhead;
        ListNode fast = dummyhead;

        while(n > 0) {
            fast = fast.next;
            n--;
        }

        while(fast.next != null) {
            fast = fast.next;
            slow = slow.next;
        }
        slow.next = slow.next.next;
        return dummyhead.next;
    }
}

一个小细节：如果不使用虚拟头节点  在删除头节点的过程中会出错 

2.判断是否带环

https://leetcode.cn/problems/linked-list-cycle/description/

代码实现：

1.快慢指针

public class Solution {
    public boolean hasCycle(ListNode head) {
        if(head == null || head.next == null) return false;

        ListNode fast = head;
        ListNode slow = head;

        // 只要有环 则fast和slow一定会相遇
        while(fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;

            if(fast == slow) return true;
        }

        return false;
        
    }
}

2.使用哈希表

public class Solution {
    public boolean hasCycle(ListNode head) {
        Set<ListNode> seen = new HashSet<ListNode>();
        while(head != null) {
            if(!seen.add(head)) {
                return true;
            }

            head = head.next;
        }

        return false;
        
    }
}

 2.环形链表的进阶

https://leetcode.cn/problems/linked-list-cycle-ii/submissions/

 

代码实现：

1.哈希表的实现
public class Solution {
    public ListNode detectCycle(ListNode head) {
        ListNode cur = head;
        Set<ListNode> seen = new HashSet<>();

        while(cur != null) {
            // 如果包含  则证明走到了入口点
            if(seen.contains(cur)) {
                return cur;
            }else {
                seen.add(cur);
            }

            cur = cur.next;
        }

        return null;
    }
}

2.双指针
public class Solution {
    public ListNode detectCycle(ListNode head) {


        // 先找相遇点
        ListNode fast = head;
        ListNode slow = head;


        while(fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;


            if(fast == slow) break;
        }


        // 跳出循环有两种condition
        if(fast == null || fast.next == null) return null;


        // fast = head  两个指针一起走 直到相同
        fast = head;
        while(fast != slow) {
            fast = fast.next;
            slow = slow.next;
        }


        return slow;
    }
}