lintcode ：Remove Duplicates from Sorted List 删除排序链表中的重复元素

题目：

删除排序链表中的重复元素

给定一个排序链表，删除所有重复的元素每个元素只留下一个。

 

您在真实的面试中是否遇到过这个题？

样例

给出1->1->2->null，返回 1->2->null

给出1->1->2->3->3->null，返回 1->2->3->null

解题：

Java程序

/**
 * Definition for ListNode
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param ListNode head is the head of the linked list
     * @return: ListNode head of linked list
     */
    public static ListNode deleteDuplicates(ListNode head) {
        // write your code here
        ListNode p = new ListNode(0);
        p.next = head;
        if(head==null)
            return null;
        while(head.next!=null){
            if(head.val==head.next.val){
                head.next = head.next.next;
            }else
                head = head.next;
        }

        return p.next;
    }
}

总耗时: 2604 ms

这个是两个比较，相同就删除一个直接删除,指针变换的比较多

可以向上面一个删除有序数组中相同的元素那样进行

head指向开始节点，current向前走，知道current.val!=head.val时候，head的指针指向current节点

Java程序：

/**
 * Definition for ListNode
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param ListNode head is the head of the linked list
     * @return: ListNode head of linked list
     */
    public static ListNode deleteDuplicates(ListNode head) {
        // write your code here
        ListNode p = new ListNode(0);
        p.next = head;
        ListNode current = new ListNode(0);
        current.next = head;
        current = current.next;
        if(head==null)
            return null;
        while(head!=null){
            while(current!=null && head.val==current.val){
                current = current.next;
            }
            head.next = current;
            head = head.next;
        }
        return p.next;
    }
}

总耗时: 1808 ms

Python程序：

"""
Definition of ListNode
class ListNode(object):
    def __init__(self, val, next=None):
        self.val = val
        self.next = next
"""
class Solution:
    """
    @param head: A ListNode
    @return: A ListNode
    """
    def deleteDuplicates(self, head):
        # write your code here
        p = ListNode(0)
        p.next = head
        if head==None:
            return None
        while head.next!=None:
            if head.val==head.next.val:
                head.next = head.next.next;
            else:
                head=head.next
        return p.next

总耗时: 319 ms

上面的第二个方法

Python程序：

"""
Definition of ListNode
class ListNode(object):
    def __init__(self, val, next=None):
        self.val = val
        self.next = next
"""
class Solution:
    """
    @param head: A ListNode
    @return: A ListNode
    """
    def deleteDuplicates(self, head):
        # write your code here
        if head==None:
            return None
        p = ListNode(0)
        p.next = head
        cur = ListNode(0)
        cur.next = head
        cur = cur.next
        while head!=None:
            while cur!=None and cur.val==head.val:
                cur = cur.next
            head.next = cur
            head = head.next
        return p.next

总耗时: 471 ms