14. 邻接链表相关简单api封装

1. 创建节点

      function CreateNode (key) {
        this.key = key;
        this.next = null;
      }
      
      let node1 = new CreateNode(1);
      let node2 = new CreateNode(2);
      let node3 = new CreateNode(3);
      let node4 = new CreateNode(4);

2. 尾插,
   我们只需要知道头节点, 就可以找到最后一个节点,
   就可以找到任何一个节点

     function insert (node,lastnode) {
          
          while (node.next){
            node = node.next;
          }
          node.next = lastnode;
     }
     insert(node1,new CreateNode(2));
     insert(node1,new CreateNode(3));
     insert(node1,new CreateNode(4));
     insert(node1,new CreateNode(5));

3. 有头节点

     let head = {type : "head",next : null};
     function insert (head,lastnode) {
          
          while (head.next){
            head = head.next;
          }
          head.next = lastnode;
     }
     insert(head,new CreateNode(2));
     insert(head,new CreateNode(3));
     insert(head,new CreateNode(4));

4. 任意位置插入

     let head = {type : "head",next : null};
     function insert (head,node,num = Infinity) {
          if (nun < 1) {
            return
          }
          while (head.next && --num > 0){
            head = head.next;
          }
          // 先临时保存
          let nextnode = head.next;
          // 插入新节点, 此时与原来的next 断开了
          head.next = node;
          // 重新接上.
          node.next = nextnode;
     }

5. 删除

根据节点删除
     function deleteNode (head,node) {
        while (head.next){
            if (head.next == node) {
                head.next = head.next.next;
                  return
            }
            head = head.next;
        }
     }

根据节点值删除
     function deleteNode (head,key) {
        while (head.next){
            if (head.next.key === key) {
                head.next = head.next.next;
                    return 此处不return 就是把链表中所有符合条件都干掉
                              此处return 就表示只干掉第一个满足条件的节点
            }
            head = head.next;
        }
     }

根据位置删除? 但实际上位置是相对来讲动态的.
     function deleteNode (head,num) {
       if (num < 1) {
        return
       }
        while (head.next && num-- > 0){
            if (num === 0) {
                head.next = head.next.next;
                return
            }
            head = head.next;
        }
     }

6. 判断有没有环

• 取一个步长为一, 取一个步长为二 的 指针
• 如果出现两个指针相同, 且不为空, 则表明形成环了.
• 好聪明, 这就像是运动场上的跑步

     function findRing (head) {
        let point1 = head.next;
        let point2 = point1 && point1.next;// head.next.next 有可能报错
        
        while (point1 && point2){ // point1 point1 都存在且不为空
            if (point1 === point2) {
                return true
            }
            point1 = point1.next;
            point2 = point2.next;
            point2 = point2 && point2.next;
        }
        return false;
     }