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             else11                 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 }