Leetcode 6

LinkedList Easy

1. 83. Remove Duplicates from Sorted List

　　当前节点值如果和下一个节点值相同直接略过。

 1 class Solution {
 2     public ListNode deleteDuplicates(ListNode head) {
 3         if( head == null)
 4             return null;
 5         ListNode node = head;
 6         while( node.next != null){
 7             if(node.next.val == node.val){
 8                 node.next = node.next.next;
 9             }
10             else
11                 node = node.next;
12         }
13         return head;
14     }
15 }

2. 141. Linked List Cycle

　　用快慢针，如果有循环快慢针会在某一时刻相同

 1 public class Solution {
 2     public boolean hasCycle(ListNode head) {
 3         if( head == null || head.next == null)
 4             return false;
 5         ListNode fast = head;
 6         ListNode slow = head;
 7         while( fast != null && fast.next != null){
 8             slow = slow.next;
 9             fast = fast.next.next;
10             if( slow == fast)
11                 return true;
12         }
13         return false;
14     }
15 }

3. 234. Palindrome Linked List

　　用快慢针找到链表重点，我们可以在找到中点后，将后半段的链表翻转一下，这样我们就可以按照回文的顺序比较了。

 1 class Solution {
 2     public boolean isPalindrome(ListNode head) {
 3         if( head == null || head.next == null)
 4             return true;
 5         ListNode fast = head;
 6         ListNode slow = head;
 7         while(fast.next !=null && fast.next.next !=null){
 8             slow = slow.next;
 9             fast = fast.next.next;
10         }
11         ListNode pre = head, last = slow.next;
12         while( last.next != null){
13             ListNode temp = last.next;
14             last.next = temp.next;
15             temp.next = slow.next;
16             slow.next = temp;
17         }
18         while(slow.next != null){
19             slow = slow.next;
20             if(pre.val != slow.val)
21                 return false;
22             pre = pre.next;
23         }
24         return true;
25     }
26 }