Given a linked list, remove the n-th node from the end of list and return its head.
Example:
Given linked list: 1->2->3->4->5, and n = 2.
After removing the second node from the end, the linked list becomes 1->2->3->5.
Note:
Given n will always be valid.
Follow up:
Could you do this in one pass?
Solution:
Find the depth of each node recursively, handle last and first node
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* next = NULL;
int tar;
int depth(ListNode* n){
if(!n){
return 0;
}
int d = depth(n->next) + 1;
if(d == tar - 1){
next = n;
}
else if(d == tar + 1){
n -> next = next;
}
return d;
}
ListNode* removeNthFromEnd(ListNode* head, int n) {
tar = n;
if(head == NULL || n <= 0) return head;
if(depth(head) == n){
return head->next;
}
return head;
}
};
Two pointer 1. advance q by n 2. advance p,q until q->next reach end, then p is n+1 to last.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode *p = head;
ListNode *q = p;
for(int i=0;i<n;i++) q = q->next;
if(q==NULL){
head = head->next;
}
else{
while(q->next!=NULL){
p=p->next;
q=q->next;
}
p->next=p->next->next;
}
return head;
}
};