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Copy path19. Remove Nth Node From End of List.cpp
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19. Remove Nth Node From End of List.cpp
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/*
Link: https://leetcode.com/problems/remove-nth-node-from-end-of-list/
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?
*/
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode* fast = head, *slow = NULL; // taking two pointer fast and slow, at the end fast will point to NULL
// and slow will point to the previous of the target node.
while(n--) { // Moving the fast pointer n nodes ahead of slow.
fast = fast -> next;
}
if (!fast) { // fast has reached null it means the head is to removed.
return head->next;
}
while(fast) { // iterating until fast become NULL
fast = fast -> next;
if(!slow) { // if slow is NULL
slow = head;
} else {
slow = slow -> next;
}
}
slow -> next = slow->next->next; // as slow points to the previous node of the target node.
// we save the next of target node to the next of the current node.
// Hence removing the current node.
return head;
}
};
/*
Time Complexity = O(L) where L is the length of the list.
Space Complexity = O(1)
*/