Coder's Cat

LeetCode: Remove Nth Node From End of List

2019-12-24

Description

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.

Naive solution

First, we need to get the length of the link list, the time complexity will be $O(N)$, we can not avoid this step, so this is also the best time complexity.

Given the length of link list, remove the n-th node will be easy if we find the node at pos length-n.

Note a dummy node will make the code cleaner.

file:img/CAPTURE-2019_12_24_remove-nth-node-from-end-of-list.org_20191224_112700.png

class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode* dummy = new ListNode(0);
dummy->next = head;
ListNode* cur = head;
int len = 0;
while(cur) {
cur = cur->next;
len++;
}
len -= n;
cur = dummy;
while(len > 0) {
cur = cur->next;
len--;
}
cur->next = cur->next->next;
return dummy->next;
}
};

Approach with two pointers

This is a clever approach, if we use two pivot pointers, their distance is n+1. Then if we move the tail forward to the end of linklist, the prev will be the pointer whose next can be deleted!

file:img/2019_12_24_remove-nth-node-from-end-of-list.org_20200204_230523.png

class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode root(0);
root.next = head;
ListNode* prev = &root;
ListNode* tail = head;

while(n--) tail = tail->next;

while (tail){
tail = tail->next;
prev = prev->next;
}

prev->next = prev->next->next;
return root.next;
}
};

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