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
java
// 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
java
// 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/
java
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
java
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
java
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:
java
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;
}
}