141. 环形链表
1.哈希表法:
将节点依次加入set，如果有重复就是有环。
public class Solution {
public boolean hasCycle(ListNode head) {
Set<ListNode> set = new HashSet<ListNode>();
while(head!=null){
if(!set.add(head)){
return true;
}
head = head.next;
}
return false;
}
}
2.快慢指针
public class Solution {
public boolean hasCycle(ListNode head) {
ListNode fast = head;
ListNode slow = head;
while(fast!=null&&fast.next!=null){
slow = slow.next;
fast = fast.next.next;
if(slow==fast){
return true;
}
}
return false;
}
}
如果有环，快慢指针迟早都会进入环，只要快指针比慢指针快1步，迟早能追上，而不会跨过。同理慢指针两步。快指针三步也可以。
//如下，快指针三步，慢指针两步。
public class Solution {
public boolean hasCycle(ListNode head) {
ListNode fast = head;
ListNode slow = head;
while(fast!=null&&fast.next!=null&&fast.next.next!=null){
slow = slow.next.next;
fast = fast.next.next.next;
if(slow==fast){
return true;
}
}
return false;
}
}
环形链表 II https://leetcode.cn/problems/linked-list-cycle-ii/solutions/12616/linked-list-cycle-ii-kuai-man-zhi-zhen-shuang-zhi-/
哈希表简单容易理解,就是空间复杂度是O(N)
/**
* 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) {
Set<ListNode> set = new HashSet<ListNode>();
ListNode t = head;
while(t!=null){
if(set.contains(t)){
return t;
}else{
set.add(t);
}
t = t.next;
}
return null;
}
}
双指针
参考大神:
作者：Krahets
链接：https://leetcode.cn/problems/linked-list-cycle-ii/solutions/12616/linked-list-cycle-ii-kuai-man-zhi-zhen-shuang-zhi-/
public class Solution {
public ListNode detectCycle(ListNode head) {
ListNode fast = head, slow = head;
while (true) {
if (fast == null || fast.next == null) return null;
fast = fast.next.next;
slow = slow.next;
if (fast == slow) break;
}
fast = head;
while (slow != fast) {
slow = slow.next;
fast = fast.next;
}
return fast;
}
}
注意：如果慢指针是两步，快指针是三步，也是可以的。相当于f=3/2s f = s+nb 消完后还是可以的，能通过测试用例
public class Solution {
public ListNode detectCycle(ListNode head) {
ListNode fast = head, slow = head;
while (true) {
if (fast == null || fast.next == null||fast.next.next == null) return null;
fast = fast.next.next.next;
slow = slow.next.next;
if (fast == slow) break;
}
fast = head;
while (slow != fast) {
slow = slow.next;
fast = fast.next;
}
return fast;
}
}