【題目描述】
Given a sortedlinked list, delete all nodes that have duplicate numbers, leaving only distinct numbersfrom the original list.
For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.
【編程步驟】
* 1. 處理特殊情況:如果該鏈表爲空或只有一個結點,說明此鏈表沒有重複的元素,直接返回頭指針head;
即:if(head==null||head.next==null)
return head;
* 2. 設置僞頭結點;
即:ListNode pre=new ListNode(0);
pre.next=head;
head=pre;
* 3. 令cur=pre.next,若不存在複本,cur依然爲pre.next;若存在複本,cur指向複本的下一個不等於它本身值的結點,並作移除操作;
ListNode cur=pre.next;
while(cur.next!=null&&cur.next.val==cur.val)
cur=cur.next;
if(pre.next==cur){ // don't find the duplicate
pre=pre.next;
}else{ //remove the duplicate node
pre.next=cur.next;
}
* 4. 返回head.next,因爲此時的head爲僞頭結點;
【代碼實現】
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode deleteDuplicates(ListNode head) {
if(head==null||head.next==null)
return head;
ListNode pre=new ListNode(0);
pre.next=head;
head=pre;
while(pre.next!=null){
ListNode cur=pre.next;
while(cur.next!=null&&cur.next.val==cur.val)
cur=cur.next;
if(pre.next==cur){ // don't find the duplicate
pre=pre.next;
}else{ //remove the duplicate node
pre.next=cur.next;
}
}
return head.next;
}
}