数据结构与算法学习day4之单向链表

1.单向链表的的定义

链表是有序的列表，这是它在内存中的存储，如下图所示：

1. 链表是以节点的形式存储的，是链式存储
2. 每个节点都包含两个部分，一个是 data 域，一个是 next 域指向下一个节点
3. 每个节点不一定是连续存储
4. 链表分为带头节点和不带头节点

2.单向链表的实现思路

（1）添加

添加节点的时候，直接添加到尾部。如下示意图：

1. 先创建一个head节点，代表单向链表的头部
2. 每添加一个节点就加入链表的后面

 （2）按指定位置添加

添加节点的时候，按照排名插入到指定位置。如下示意图：

1. 通过遍历找到添加指针的位置，需要temp临时指针标记
2. 新节点.next = temp.next
3. temp.next = 新节点

（3）修改

1. 通过遍历找到节点

2. 创建要修改的新节点对象

3. 将新节点的属性复制给节点

4. 删除节点 

（4）删除

1. 通过遍历找到需要删除节点的前一个节点，用temp标记
2. temp.next = temp.next.next
3. 被删除的节点会有java的垃圾回收机制处理

3.代码实现

package DataStructures.linkedlist;

import java.util.Stack;

public class SingleLinkedListDemo {

	public static void main(String[] args) {
		HeroNode hero1 = new HeroNode(1, "宋江", "及时雨");
		HeroNode hero2 = new HeroNode(2, "卢俊义", "玉麒麟");
		HeroNode hero3 = new HeroNode(3, "吴用", "智多星");
		HeroNode hero4 = new HeroNode(4, "林冲", "豹子头");

		SingleLinkedList singleLinkedList = new SingleLinkedList();
		

		singleLinkedList.add(hero1);
		singleLinkedList.add(hero4);
		singleLinkedList.add(hero2);
		singleLinkedList.add(hero3);

		singleLinkedList.list();

		reversePrint(singleLinkedList.getHead());
		
/*
		singleLinkedList.addByOrder(hero1);
		singleLinkedList.addByOrder(hero4);
		singleLinkedList.addByOrder(hero2);
		singleLinkedList.addByOrder(hero3);

		singleLinkedList.list();

		HeroNode newHeroNode = new HeroNode(2, "", "~~");
		singleLinkedList.update(newHeroNode);

		singleLinkedList.list();

		singleLinkedList.del(1);
		singleLinkedList.del(4);
		singleLinkedList.list();
		

		System.out.println(getLength(singleLinkedList.getHead()));//2
		

		HeroNode res = findLastIndexNode(singleLinkedList.getHead(), 3);
		System.out.println("res=" + res);
*/		
		
	}

	public static void reversePrint(HeroNode head) {
		if(head.next == null) {
			return;
		}
		Stack<HeroNode> stack = new Stack<HeroNode>();
		HeroNode cur = head.next;

		while(cur != null) {
			stack.push(cur);
			cur = cur.next;
		}

		while (stack.size() > 0) {
			System.out.println(stack.pop());
		}
	}

	public static void reversetList(HeroNode head) {
		if(head.next == null || head.next.next == null) {
			return ;
		}

		HeroNode cur = head.next;
		HeroNode next = null;
		HeroNode reverseHead = new HeroNode(0, "", "");

		while(cur != null) { 
			next = cur.next;
			cur.next = reverseHead.next;
			reverseHead.next = cur;
			cur = next;
		}
		head.next = reverseHead.next;
	}

	public static HeroNode findLastIndexNode(HeroNode head, int index) {
		if(head.next == null) {
			return null;
		}

		int size = getLength(head);
		if(index <=0 || index > size) {
			return null; 
		}
		HeroNode cur = head.next;
		for(int i =0; i< size - index; i++) {
			cur = cur.next;
		}
		return cur;
		
	}

	public static int getLength(HeroNode head) {
		if(head.next == null) {
			return 0;
		}
		int length = 0;
		HeroNode cur = head.next;
		while(cur != null) {
			length++;
			cur = cur.next;
		}
		return length;
	}

}

class SingleLinkedList {
	private HeroNode head = new HeroNode(0, "", "");
	

	public HeroNode getHead() {
		return head;
	}

	public void add(HeroNode heroNode) {

		HeroNode temp = head;
		while(true) {
			if(temp.next == null) {
				break;
			}
			temp = temp.next;
		}

		temp.next = heroNode;
	}

	public void addByOrder(HeroNode heroNode) {

		HeroNode temp = head;
		boolean flag = false;
		while(true) {
			if(temp.next == null) {
				break;
			} 
			if(temp.next.no > heroNode.no) {
				break;
			} else if (temp.next.no == heroNode.no) {
				
				flag = true;
				break;
			}
			temp = temp.next;
		}

		if(flag) {
			System.out.print(heroNode.no);
		} else {
			heroNode.next = temp.next;
			temp.next = heroNode;
		}
	}

	public void update(HeroNode newHeroNode) {

		if(head.next == null) {
			System.out.println("链表为空");
			return;
		}

		HeroNode temp = head.next;
		boolean flag = false;
		while(true) {
			if (temp == null) {
				break;
			}
			if(temp.no == newHeroNode.no) {
				flag = true;
				break;
			}
			temp = temp.next;
		}
		if(flag) {
			temp.name = newHeroNode.name;
			temp.nickname = newHeroNode.nickname;
		} else {
			System.out.printf("没有找到编号 %d 的节点\n", newHeroNode.no);
		}
	}
	

	public void del(int no) {
		HeroNode temp = head;
		boolean flag = false;
		while(true) {
			if(temp.next == null) {
				break;
			}
			if(temp.next.no == no) {
				flag = true;
				break;
			}
			temp = temp.next;
		}
		if(flag) {
			temp.next = temp.next.next;
		}else {
			System.out.printf("要删除的%d节点不存在\n", no);
		}
	}

	public void list() {
		if(head.next == null) {
			System.out.println("链表为空");
			return;
		}
		HeroNode temp = head.next;
		while(true) {
			if(temp == null) {
				break;
			}
			System.out.println(temp);
			temp = temp.next;
		}
	}
}

class HeroNode {
	public int no;
	public String name;
	public String nickname;
	public HeroNode next;
	public HeroNode(int no, String name, String nickname) {
		this.no = no;
		this.name = name;
		this.nickname = nickname;
	}

	@Override
	public String toString() {
		return "HeroNode [no=" + no + ", name=" + name + ", nickname=" + nickname + "]";
	}
	
}

 4.总结

单向链表的实现还是比较简单的，最主要的是要弄清出每个节点怎么实现连接，删除节点如何操作。