LeetCode 203. Remove Linked List Elements
2016-07-26 16:57
330 查看
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
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* removeElements(ListNode* head, int val) {
if(!head || !head->next && head->val == val) return NULL;
ListNode* pre;
ListNode *node;
ListNode *tmp;
pre = node = head;
while(head && head->val == val){
node = head;
head = head -> next;
delete node;
}
if(!head) return NULL;
pre = head;
node = head -> next;
while(node){
if(node->val == val){
pre -> next = node -> next;
delete node;
node = pre -> next;
}else{
node = node -> next;
pre = pre -> next;
}
}
return head;
}
};
Example
Given: 1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6, val = 6
Return: 1 --> 2 --> 3 --> 4 --> 5
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* removeElements(ListNode* head, int val) {
if(!head || !head->next && head->val == val) return NULL;
ListNode* pre;
ListNode *node;
ListNode *tmp;
pre = node = head;
while(head && head->val == val){
node = head;
head = head -> next;
delete node;
}
if(!head) return NULL;
pre = head;
node = head -> next;
while(node){
if(node->val == val){
pre -> next = node -> next;
delete node;
node = pre -> next;
}else{
node = node -> next;
pre = pre -> next;
}
}
return head;
}
};
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