Remove all elements from a linked list of integers that have value val.
Example
- Given:
1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6,val = 6 - Return:
1 --> 2 --> 3 --> 4 --> 5
C Solution 1:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* removeElements(struct ListNode* head, int val) {
struct ListNode **p = &head;
while (*p) {
if ((*p)->val == val) {
struct ListNode *tmp = (*p)->next;
free(*p);
*p = (*p)->next;
}
else p = &((*p)->next);
}
return head;
}
C Solution 2:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* removeElements(struct ListNode* head, int val) {
struct ListNode dummy, *p = &dummy;
p->next = head;
while (p->next) {
if (p->next->val == val) {
head = p->next->next;
free(p->next);
p->next = head;
}
else p = p->next;
}
return dummy.next;
}
Python Solution:
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def removeElements(self, head, val):
"""
:type head: ListNode
:type val: int
:rtype: ListNode
"""
dummy = ListNode(0)
dummy.next = head
cur = dummy
while cur.next:
if cur.next.val == val:
cur.next = cur.next.next
else:
cur = cur.next
return dummy.next
Summary:
- 9ms, 25.69%
- If you need to find a node and then delete it, double pointer is what you want. Of course, you can use pre pointer, but that's not beautiful.
LeetCode: 203. Remove Linked List Elements
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