剑指offer第二版-18.2删除排序链表中重复的节点

本系列导航:剑指offer(第二版)java实现导航帖

面试题18题目二:删除排序链表中重复的节点

题目要求:
比如[1,2,2,3,3,3],删除之后为[1];

解题思路:
由于是已经排序好的链表,需要确定重复区域的长度,删除后还需要将被删去的前与后连接,所以需要三个节点pre,cur,post,cur-post为重复区域,删除后将pre与post.next连接即可。
此外,要注意被删结点区域处在链表头部的情况,因为需要修改head。

package structure;
/**
 * Created by ryder on 2017/6/13.
 */
public class ListNode<T> {
    public T val;
    public ListNode<T> next;
    public ListNode(T val){
        this.val = val;
        this.next = null;
    }
    @Override
    public String toString() {
        StringBuilder ret = new StringBuilder();
        ret.append("[");
        for(ListNode cur = this;;cur=cur.next){
            if(cur==null){
                ret.deleteCharAt(ret.lastIndexOf(" "));
                ret.deleteCharAt(ret.lastIndexOf(","));
                break;
            }
            ret.append(cur.val);
            ret.append(", ");
        }
        ret.append("]");
        return ret.toString();
    }
}
package chapter3;

import structure.ListNode;

/**
 * Created by ryder on 2017/7/7.
 * 删除排序链表中的重复节点
 */
public class P122_deleteDuplicatedNode {
    public static ListNode<Integer> deleteDuplication(ListNode<Integer> head){
        if(head==null||head.next==null)
            return head;
        ListNode<Integer> pre = null;
        ListNode<Integer> cur = head;
        ListNode<Integer> post = head.next;
        boolean needDelete = false;
        while (post!=null){
            if(cur.val.equals(post.val)){
                needDelete = true;
                post=post.next;
            }
            else if(needDelete && !cur.val.equals(post.val)){
                if(pre==null)
                    head = post;
                else
                    pre.next=post;
                cur = post;
                post = post.next;
                needDelete = false;
            }
            else{
                pre = cur;
                cur = post;
                post = post.next;
            }
        }
        if(needDelete && pre!=null)
            pre.next = null;
        else if(needDelete && pre==null)
            head = null;
        return head;
    }
    public static void main(String[] args){
        ListNode<Integer> head = new ListNode<>(1);
        head.next= new ListNode<>(1);
        head.next.next = new ListNode<>(2);
        head.next.next.next = new ListNode<>(2);
        head.next.next.next.next = new ListNode<>(2);
        head.next.next.next.next.next = new ListNode<>(3);
        System.out.println(head);
        head = deleteDuplication(head);
        System.out.println(head);
    }
}

运行结果

[1, 2, 3]
[1, 2]
[2]
    原文作者:ryderchan
    原文地址: https://www.jianshu.com/p/efb520176364#comments
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