回文链表空间复杂度O(1)解法

src/main/java/com/cool/hot100/linkedlist/Num234.java

@@ -1,5 +1,7 @@
 package com.cool.hot100.linkedlist;
 
+import org.junit.Test;
+
 public class Num234 {
 
     /**
@@ -12,22 +14,60 @@ public class Num234 {
     public ListNode node;
 
     public boolean isPalindrome(ListNode head) {
-        node = head;
+//        node = head;
-        return reNodeCheck(head);
+//        return reNodeCheck(head);
-    }
+        //时间复杂度为O(n),空间复杂度为O(1)解法
+        ListNode slowNode = head;
+        ListNode fastNode = head;
+        while (fastNode.next != null && fastNode.next.next != null) {
+            fastNode = fastNode.next.next;
+            slowNode = slowNode.next;
+        }
+        ListNode temp = head;
+        ListNode reverse = reverse(slowNode.next);
+        while (reverse != null) {
+            if (reverse.val != temp.val) {
+                return false;
+            }
+            temp = temp.next;
+            reverse = reverse.next;
 
-    //递归反向对比
-    public boolean reNodeCheck(ListNode node){
-        if (node!=null){
-            if(!reNodeCheck(node.next)){
-                return false;
-            }
-            if(node.val!=this.node.val){
-                return false;
-            }
-            this.node=this.node.next;
         }
         return true;
     }
 
+    public ListNode reverse(ListNode node) {
+        ListNode pre = null;
+        while (node != null) {
+            ListNode next = node.next;
+            node.next = pre;
+            pre = node;
+            node = next;
+        }
+        return pre;
+    }
+
+    //递归反向对比
+    public boolean reNodeCheck(ListNode node) {
+        if (node != null) {
+            if (!reNodeCheck(node.next)) {
+                return false;
+            }
+            if (node.val != this.node.val) {
+                return false;
+            }
+            this.node = this.node.next;
+        }
+        return true;
+    }
+    @Test
+    public void test(){
+//        ListNode node=new ListNode(5,null);
+        ListNode node1=new ListNode(1,null);
+        ListNode node2=new ListNode(2,node1);
+        ListNode node3=new ListNode(2,node2);
+        ListNode node4=new ListNode(1,node3);
+        isPalindrome(node4);
+    }
+
 }