Given a sorted linked list, delete all duplicates such that each element appear only once.
For example,
Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3.
總算自己做出來一道了额各，雖然只是一道easy, 但好像連續(xù)幾道都是看的答案。這道題確實簡單吧，思路都很直接追他，注意一下循環(huán)條件就好了苛蒲。（我是報錯NullPointerException后自己加的curt.next != null && curt.val == curt.next.val. 入股不加上curt.next != null, 那么curt.next.val就會報空指針颜及。）
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode deleteDuplicates(ListNode head) {
if (head == null){
return head;
}
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode curt = head;
while (curt != null && curt.next != null){
while (curt.next != null && curt.val == curt.next.val){
curt.next = curt.next.next;
}
curt = curt.next;
}
return dummy.next;
}
}
