代码随想录打卡 03

关于链表这种数据结构相信大家都十分熟悉，谈起来都能如数家珍，但是如果让你写链表相关的算法题的时候就很容易出错，处理不好链表循环中的引用的问题。这个时候需要将每个节点当作一个对象来看待，这个对象中有个next 引用，指向的下一个节点。同时对象中保存了当前节点的值，一旦修改了next引用，那么这个节点的下一个节点就指向了其他的节点，也就意味着这个链表发生了变化。

删除链表中的元素

leetcode 202

在处理头节点和其他节点的时候需要做下区别，这里使用了一个虚拟的头节点

class Solution {
    public ListNode removeElements(ListNode head, int val) {

        ListNode dummy = new ListNode(-1, head);//定义一个虚拟的头节点
        ListNode prev = dummy;
        ListNode current = head;

        while(current != null) {
            if(current.val == val) {
              prev.next = current.next;
            } else {
                prev = current;
            }
             current = current.next;

        }
        return dummy.next;

    }
}

设计一个链表

设计一个链表，包含一些简单的操作。

leetcode 707

class MyLinkedList {

  ListNode head;

  class ListNode {
    int val;
    ListNode next;
    ListNode() {}
    ListNode(int val) { this.val = val; }
    ListNode(int val, ListNode next) { this.val = val; this.next = next; }
  }

  public MyLinkedList() {

  }

  public int get(int index) {
    ListNode current = head;
    int pos = index;
    int result = -1;
    while(current != null && pos >= 0) {
      if(pos == 0) {
        result =  current.val;
        break;
      }
      current = current.next;
      pos --;
    }
    return result;
  }

  public void addAtHead(int val) {
    ListNode newHead = new ListNode(val, this.head);
    this.head = newHead;

  }

  public void addAtTail(int val) {
    if (head == null) {
      addAtHead(val);
      return;
    }
    ListNode current = head;
    while(current.next != null) {
      current = current.next;
    }

    current.next = new ListNode(val, null);

  }

  public void addAtIndex(int index, int val) {
    int pos = index - 1;

    ListNode current = head;

    if(index <= 0) {
      head = new ListNode(val, head);
    }

    while(current != null && pos >= 0) {
      if(pos == 0) {
        current.next = new ListNode(val, current.next);
        break;
      }
      current = current.next;
      pos --;
    }
  }

  public void deleteAtIndex(int index) {

    ListNode current = head;
    int pos = index - 1;

    if(index == 0) {
      this.head = head.next;
    }

    while(current != null && current.next != null && pos >= 0) {
      if(pos == 0) {
        current.next = current.next.next;
        break;
      }
      current = current.next;
      pos --;
    }
  }

  @Override
  public String toString() {
    ListNode current = head;
    StringBuilder sb = new StringBuilder();
    while (current != null) {
      sb.append(current.val);
      sb.append(",");
      current = current.next;
    }
    return sb.toString();
  }
}

反转链表

leetcode 206

经典的链表相关的题目，之前也是做过很多遍，再做的时候依然修修改改花了些时间

class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode current = head;
        ListNode prev = null;
        // ListNode result = head;
        while(current != null) {
            ListNode next = current.next;
            current.next = prev;
            prev = current;
            current = next;
        }
        return prev;

    }
}