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
.
思路:
简单题,没什么好说的。
class Solution {
public:
ListNode *deleteDuplicates(ListNode *head) {
if(head == NULL)
return NULL; ListNode * ans = head;
ListNode * anscur = ans; //记录答案链表的最后一个指针
ListNode * cur = head->next; //记录判断的是原链表中的哪一个指针
while(cur != NULL)
{
if(anscur->val != cur->val) //只有值变化的时候才把指针接到ans链表后面 节省操作
{
anscur->next = cur;
anscur = anscur->next;
}
cur = cur->next;
}
anscur->next = NULL; //防止最后的指针是重复的,给ans结尾
return ans;
}
};
大神的代码更加简洁, 没有必要新建一个头结点,就用原来的头结点就好。
class Solution {
public:
ListNode *deleteDuplicates(ListNode *head) {
if(NULL == head) return head; ListNode *cur = head, *nxt = head->next;
while (nxt) {
if (cur->val != nxt->val)
cur = cur->next = nxt;
else if (!nxt->next)
cur->next = nxt->next;
nxt = nxt->next;
} return head;
}
};