LeetCode707.设计链表

题目来源：https://leetcode-cn.com/problems/design-linked-list/description/

题目描述：

代码如下：

class MyLinkedList {

	int length;
	Node head;

	class Node {
		int val;
		Node next;

		Node(int x) {
			this.val = x;
		}
	}

	public MyLinkedList() {
		this.length = 0;
		this.head = null;
	}

	public int get(int index) {
		if (index < 0 || index >= this.length) {
			return -1;
		}
		int count = 0;
		Node curr = head;
		while (count < index) {
			curr = curr.next;
			count++;
		}
		return curr.val;
	}

	public void addAtHead(int val) {
		Node newNode = new Node(val);// 初始化新节点
		newNode.next = this.head;// 新节点的下一个结点为原来的头结点
		this.head = newNode;// 将新节点设置成头结点
		this.length++;// 长度+1
	}

	public void addAtTail(int val) {
		if (this.length == 0) {
			head = new Node(val);
			return;
		}
		Node newNode = new Node(val);
		Node temp = head;
		while (temp.next != null) {
			temp = temp.next;
		}
		temp.next = newNode;
		this.length++;
	}

	public void addAtIndex(int index, int val) {
		if (index == this.length) {// 尾结点的下一个结点
			addAtTail(val);
			return;
		}
		if (index > this.length) {// 越界
			return;
		}
		Node newNode = new Node(val);
		Node temp = head;
		int counter = 0;
		while (counter < (index - 1)) {// 在原有链表里
			temp = temp.next;
			counter++;
		}
		newNode.next = temp.next;
		temp.next = newNode;
		this.length++;
	}

	public void deleteAtIndex(int index) {
		if (index < 0 || index >= this.length) {
			return;
		}
		// 要删除头结点，直接让head指向下一个结点，GC会回收
		if (index == 0) {
			head = head.next;
		}
		Node curr = head; // 用于保存被删结点
		Node pre = null; // 用于保存被删结点的前驱结点
		int count = 0;
		while (count < index) {
			pre = curr;
			curr = curr.next;
			count++;
		}
		// 让前驱结点执行被删结点的后继结点，完成结点删除
		pre.next = curr.next;
		this.length--;
	}
}