707.Design-Linked-List
707. Design Linked List
题目地址
https://leetcode.com/problems/design-linked-list/
题目描述
Design your implementation of the linked list. You can choose to use the singly linked list or the doubly linked list. A node in a singly linked list should have two attributes: val and next. val is the value of the current node, and next is a pointer/reference to the next node. If you want to use the doubly linked list, you will need one more attribute prev to indicate the previous node in the linked list. Assume all nodes in the linked list are 0-indexed.
Implement these functions in your linked list class:
get(index) : Get the value of the index-th node in the linked list. If the index is invalid, return -1.
addAtHead(val) : Add a node of value val before the first element of the linked list. After the insertion, the new node will be the first node of the linked list.
addAtTail(val) : Append a node of value val to the last element of the linked list.
addAtIndex(index, val) : Add a node of value val before the index-th node in the linked list. If index equals to the length of linked list, the node will be appended to the end of linked list. If index is greater than the length, the node will not be inserted.
deleteAtIndex(index) : Delete the index-th node in the linked list, if the index is valid.
Example:
Input:
["MyLinkedList","addAtHead","addAtTail","addAtIndex","get","deleteAtIndex","get"]
[[],[1],[3],[1,2],[1],[1],[1]]
Output:
[null,null,null,null,2,null,3]
Explanation:
MyLinkedList linkedList = new MyLinkedList(); // Initialize empty LinkedList
linkedList.addAtHead(1);
linkedList.addAtTail(3);
linkedList.addAtIndex(1, 2); // linked list becomes 1->2->3
linkedList.get(1); // returns 2
linkedList.deleteAtIndex(1); // now the linked list is 1->3
linkedList.get(1); // returns 3
Constraints:
0 <= index,val <= 1000
Please do not use the built-in LinkedList library.
At most 2000 calls will be made to get, addAtHead, addAtTail, addAtIndex and deleteAtIndex.代码
Approach #1 Singly Linked List
Complexity Analysis
Time complexity: O(1) for addAtHead. O(k) for get, addAtIndex, and deleteAtIndex, where k is an index of the element to get, add or delete. O(N) for addAtTail.
Space complexity: O(1) for all operations.
class ListNode {
int val;
ListNode next;
ListNode(int x) {val = x;};
}
class MyLinkedList {
int size;
ListNode head;
/** Initialize your data structure here. */
public MyLinkedList() {
size = 0;
head = new ListNode(0);
}
/** Get the value of the index-th node in the linked list. If the index is invalid, return -1. */
public int get(int index) {
if (index < 0 || index >= size) return -1;
ListNode curr = head;
for (int i = 0; i < index + 1; i++) {
curr = curr.next;
}
return curr.val;
}
/** Add a node of value val before the first element of the linked list. After the insertion, the new node will be the first node of the linked list. */
public void addAtHead(int val) {
addAtIndex(0, val);
}
/** Append a node of value val to the last element of the linked list. */
public void addAtTail(int val) {
addAtIndex(size, val);
}
/** Add a node of value val before the index-th node in the linked list. If index equals to the length of linked list, the node will be appended to the end of linked list. If index is greater than the length, the node will not be inserted. */
public void addAtIndex(int index, int val) {
if (index > size) return;
if (index < 0) return 0;
size++;
ListNode pred = head;
for (int i = 0; i < index; i++) {
pred = pred.next;
}
ListNode toAdd = new ListNode(val);
toAdd.next = pred.next;
pred.next = toAdd;
}
/** Delete the index-th node in the linked list, if the index is valid. */
public void deleteAtIndex(int index) {
if (index < 0 || index >= size) return;
size--;
ListNode pred = head;
for (int i = 0; i < index; i++) {
pred = pred.next;
}
pred.next = pred.next.next;
}
}
/**
* Your MyLinkedList object will be instantiated and called as such:
* MyLinkedList obj = new MyLinkedList();
* int param_1 = obj.get(index);
* obj.addAtHead(val);
* obj.addAtTail(val);
* obj.addAtIndex(index,val);
* obj.deleteAtIndex(index);
*/Approach #2 Doubly Linked List
Complexity Analysis
Time complexity: O(1) for addAtHead and addAtTail. O(min(k,N−k)) for get, addAtIndex, and deleteAtIndex, where kk is an index of the element to get, add or delete.
Space complexity: O(1) for all operations.
class ListNode {
int val;
ListNode next;
ListNode prev;
ListNode(int x) { val = x; }
}
class MyLinkedList {
int size;
ListNode head, tail;
public MyLinkedList() {
size = 0;
head = new ListNode(0);
tail = new ListNode(0);
head.next = tail;
tail.prev = head;
}
public int get(int index) {
if (index < 0 || index >= size) return -1;
ListNode curr = head;
if (index + 1 < size - index) {
for (int i = 0; i< index + 1; i++) {
curr = curr.next;
}
} else {
curr = tail;
for (int i = 0; i < size - index; i++) {
curr = curr.prev;
}
}
return curr.val;
}
public void addAtHead(int val) {
ListNode pred = head, succ = head.next;
size++;
ListNode toAdd = new ListNode(val);
toAdd.prev = pred;
toAdd.next = succ;
pred.next = toAdd;
succ.prev = toAdd;
}
public void addAtTail(int val) {
ListNode succ = tail, pred = tail.prev;
size++;
ListNode toAdd = new ListNode(val);
toAdd.prev = pred;
toAdd.next = succ;
pred.next = toAdd;
succ.prev = toAdd;
}
public void addAtIndex(int index, int val) {
if (index > size) return;
if (index < 0) index = 0;
ListNode pred, succ;
if (index < size - index) {
pred = head;
for (int i = 0; i < index; i++) {
pred = pred.next;
}
succ = pred.next;
} else {
succ = tail;
for (int i = 0; i < size - index; i++) {
succ = succ.prev;
}
}
size++;
ListNode toAdd = new ListNode(val);
toAdd.prev = pred;
toAdd.next = succ;
pred.next = toAdd;
succ.prev = toAdd;
}
public void deleteAtIndex(int index) {
if (index < 0 || index >= size) return;
ListNode pred, succ;
if (index < size - index) {
pred = head;
for (int i = 0; i < index; i++) {
pred = pred.next;
}
succ = pred.next.next;
} else {
succ = tail;
for (int i = 0; i < size - index - 1; i++) {
succ = succ.prev;
}
pred = succ.prev.prev;
}
size--;
pred.next = succ;
succ.prev = pred;
}
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