花花酱 LeetCode 19. Remove Nth Node From End of List

By zxi on September 19, 2018

Problem

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 0: Cheating! store the nodes in an array

C++

// Author: Huahua

class Solution {

public:

ListNode *removeNthFromEnd(ListNode *head, int n) {

if (!head) return nullptr;

vector<ListNode *> nodes;

ListNode *cur = head;

while (cur) {

nodes.push_back(cur);

cur = cur->next;

}

if (n == nodes.size()) return head->next;

ListNode* nodes_to_remove = nodes[nodes.size()-n];

ListNode* parent = nodes[nodes.size() - n - 1];

parent->next = nodes_to_remove->next;

delete nodes_to_remove;

return head;

}

};

Solution 1: Two passes

Time complexity: O(L)

Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

ListNode* removeNthFromEnd(ListNode* head, int n) {

int l = 0;

ListNode* cur = head;

while (cur) {

++l;

cur = cur->next;

}

if (n == l) {

ListNode* ans = head->next;

delete head;

return ans;

}

l -= n;

cur = head;

while (--l) cur = cur->next;

ListNode* node = cur->next;

cur->next = node->next;

delete node;

return head;

}

};

Solution 2: Fast/Slow Pointers + Dummy Head / Prev

Fast pointer moves n steps first, and then slow pointer starts moving.

When fast pointer reaches tail, slow pointer is n-th node from the end.

Time complexity: O(L)

Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

ListNode* removeNthFromEnd(ListNode* head, int n) {

ListNode* fast = head;

for (int i = 0; i < n; ++i)

fast = fast->next;

ListNode dummy(0);

dummy.next = head;

ListNode* prev = &dummy;

while (fast) {

fast = fast->next;

prev = prev->next;

}

ListNode* node = prev->next;

prev->next = node->next;

delete node;

return dummy.next;

}

};

Java

// Author: Huahua

class Solution {

public ListNode removeNthFromEnd(ListNode head, int n) {

ListNode dummy = new ListNode(0);

dummy.next = head;

ListNode fast = head;

while (n-- > 0) fast = fast.next;

ListNode prev = dummy;

while (fast != null) {

fast = fast.next;

prev = prev.next;

}

prev.next = prev.next.next;

return dummy.next;

}

Python3

# Author: Huahua

class Solution:

def removeNthFromEnd(self, head, n):

dummy = ListNode(0)

dummy.next = head

fast, prev = head, dummy

for _ in range(n):

fast = fast.next

while fast:

fast, prev = fast.next, prev.next

prev.next = prev.next.next

return dummy.next

花花酱 LeetCode 328. Odd Even Linked List

By zxi on April 12, 2018

Problem

题目大意：给你一个链表，把所有奇数位置的节点串在一起，后面跟着所有串在一起的偶数位置的节点。

Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.

You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.

Example:
Given 1->2->3->4->5->NULL,
return 1->3->5->2->4->NULL.

Note:
The relative order inside both the even and odd groups should remain as it was in the input.
The first node is considered odd, the second node even and so on …

Credits:
Special thanks to @DjangoUnchained for adding this problem and creating all test cases.

Solution

Time complexity: O(n)

Space complexity: O(1)

C++

// Author: Huahua

// Running time: 21 ms

class Solution {

public:

ListNode* oddEvenList(ListNode* head) {

if (head == nullptr) return head;

ListNode dummy_odd(0);

ListNode dummy_even(0);

ListNode* prev_odd = &dummy_odd;

ListNode* prev_even = &dummy_even;

int index = 0;

while (head) {

auto next = head->next;

head->next = nullptr; // important

if (index++ & 1) {

prev_even->next = head;

prev_even = head;

} else {

prev_odd->next = head;

prev_odd = head;

}

head = next;

}

prev_odd->next = dummy_even.next;

return dummy_odd.next;

}

};

// Author: Huahua

// Running time: 18 ms

class Solution {

public:

ListNode* oddEvenList(ListNode* head) {

if (head == nullptr) return head;

vector<ListNode> heads(2, ListNode(0));

vector<ListNode*> prevs{&heads[0], &heads[1]};

int index = 0;

while (head) {

auto next = head->next;

head->next = nullptr; // important

prevs[index]->next = head;

prevs[index] = head;

head = next;

index ^= 1;

}

prevs[0]->next = heads[1].next;

return heads[0].next;

}

};

Java

// Author: Huahua

// Running time: 1 ms

class Solution {

public ListNode oddEvenList(ListNode head) {

ListNode[] heads = new ListNode[]{new ListNode(0), new ListNode(0)};

ListNode[] prevs = new ListNode[]{heads[0], heads[1]};

int index = 0;

while (head != null) {

ListNode next = head.next;

head.next = null;

prevs[index].next = head;

prevs[index] = prevs[index].next;

head = next;

index ^= 1;

}

prevs[0].next = heads[1].next;

return heads[0].next;

}

Python3

"""

Author: Huahua

Running time: 56 ms

"""

class Solution:

def oddEvenList(self, head):

heads = [ListNode(0), ListNode(0)]

prevs = [heads[0], heads[1]]

index = 0

while head:

nxt = head.next

head.next = None

prevs[index].next = head

prevs[index] = prevs[index].next

head = nxt

index ^= 1

prevs[0].next = heads[1].next

return heads[0].next

Posts tagged as “dummy head”

花花酱 LeetCode 1669. Merge In Between Linked Lists

Solution: List Operations

C++

花花酱 LeetCode 19. Remove Nth Node From End of List

Problem

Solution 0: Cheating! store the nodes in an array

C++

Solution 1: Two passes

C++

Solution 2: Fast/Slow Pointers + Dummy Head / Prev

C++

Java

Python3

花花酱 LeetCode 328. Odd Even Linked List

Problem

Solution