LeetCode 707. Design Linked List (设计链表)

题目标签:Linked List

  题目让我们自己设计一个 linked list,可以是单向和双向的。这里选的是单向,题目并不是很难,但要考虑到所有的情况,具体看code。

Java Solution:

Runtime:  56 ms, faster than 21.21% 

Memory Usage: 45.2 MB, less than 88.89%

完成日期:07/08/2019

关键点:edge cases

public class ListNode {
      int val;
      ListNode next;
    
      ListNode(int x) { 
          val = x;
      }
}

class MyLinkedList {
    
    ListNode dummyHead;
    /** Initialize your data structure here. */
    public MyLinkedList() {
        dummyHead = new ListNode(0);
        dummyHead.next = null;
    }
    
    /** 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)
            return -1;
        
        ListNode cursor = dummyHead.next;
        int i = 0;
        
        while(cursor != null) {
            if(i == index) 
                return cursor.val;
            
            i++;
            cursor = cursor.next;
        }
        
        return -1;
    }
    
    /** 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) {
        // case 1: no first node
        if(dummyHead.next == null) {
            ListNode node = new ListNode(val);
            dummyHead.next = node;
            node.next = null;
        }
        else {
            ListNode node = new ListNode(val);
            node.next = dummyHead.next;
            dummyHead.next = node;
        }
    }
    
    /** Append a node of value val to the last element of the linked list. */
    public void addAtTail(int val) {
        ListNode cursor = dummyHead;
        
        // find the node's next is null, insert node after that node
        while(cursor.next != null) { 
            cursor = cursor.next;
        }
        
        ListNode node = new ListNode(val);
        node.next = cursor.next;
        cursor.next = node;
    }
    
    /** 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 < 0) {
            addAtHead(val);
            return;
        }
            
        ListNode cursor = dummyHead;
        ListNode prevNode = dummyHead;
        int i = -1;
        
        while(cursor != null) {
            // if find the spot to insert node
            if(i + 1 == index) {
                ListNode node = new ListNode(val);
                node.next = cursor.next;
                cursor.next = node;
                
                return;
            }
            
            i++;
            cursor = cursor.next;
        }
    }
    
    /** Delete the index-th node in the linked list, if the index is valid. */
    public void deleteAtIndex(int index) {
        if(index < 0)
            return;
        
        ListNode cursor = dummyHead.next;
        ListNode prevNode = dummyHead;
        int i = 0;
        
        while(cursor != null) {
            if(i == index) { // find the node to delete
                prevNode.next = cursor.next;
                cursor.next = null;
                
                return;
            }

            i++;
            prevNode = cursor;
            cursor = cursor.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);
 */

参考资料:n/a

LeetCode 题目列表 - LeetCode Questions List

题目来源:https://leetcode.com/

原文地址:https://www.cnblogs.com/jimmycheng/p/11489391.html