2016年12月11日 星期日

[LeetCode] 82. Remove Duplicates from Sorted List II

轉自LeetCode

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from 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.
<Solution>

這題是 Remove Duplicates from Sorted List 的衍生題

是要把所有重複的 node 都拿掉,而不是只留一個

還是在考 link list 的操作,這題因為會有刪除所有 node 的情況出現

用一個 dummy head 配合兩個指針,會比較好寫

code 如下
c++
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* deleteDuplicates(ListNode* head) {
if(!head || !head->next) {
return head;
}
ListNode *dummy = new ListNode(0);
dummy->next = head;
ListNode *prv = dummy;
ListNode *cur = NULL;
while(prv->next) {
cur = prv->next;
while(cur->next && cur->next->val == cur->val) {
cur = cur->next;
}
if(cur != prv->next) {
prv->next = cur->next;
}
else {
prv = prv->next;
}
}
return dummy->next;
}
};

kotlin
/**
* Example:
* var li = ListNode(5)
* var v = li.`val`
* Definition for singly-linked list.
* class ListNode(var `val`: Int) {
* var next: ListNode? = null
* }
*/
class Solution {
fun deleteDuplicates(head: ListNode?): ListNode? {
return if(head == null || head.next == null) {
head
} else {
val dummy = ListNode(0)
dummy.next = head
var prv: ListNode? = dummy
var cur: ListNode? = null
while(prv?.next != null) {
cur = prv!!.next
while(cur?.next != null && cur!!.`val` == cur!!.next.`val`) {
cur = cur.next
}
if(cur != prv!!.next) {
prv!!.next = cur?.next
} else {
prv = prv?.next
}
}
dummy.next
}
}
}

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