Remove Duplicates from Sorted List II

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Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    public ListNode deleteDuplicates(ListNode head) {        if(head==null||head.next==null){／／其实该if语句块是直接在else里面包含的            return head;        }else{            ListNode dummy=new ListNode(0);            dummy.next=head;            ListNode preNode=dummy;            ListNode curNode=head;            while(curNode!=null){                if(curNode.next!=null){                    if(curNode.next.val!=curNode.val){                        preNode=curNode;                        curNode=curNode.next;                    }else{                        curNode=curNode.next;                        while((curNode.next!=null)&&(curNode.next.val==curNode.val)){                            curNode=curNode.next;                        }                        if(curNode==null){                            preNode.next=null;                            return dummy.next;                        }                        preNode.next=curNode.next;／／接下来两行代码，先写哪一行 效率也是不一样的，可以看下面的优化                        curNode=curNode.next;                    }                 }else{                     return dummy.next;                 }                       }           return dummy.next;                 }    }}

1）边界问题，1 刚开始便是duplicates 2 结尾是重复

2）linkedlist 利用dummy node 可以方便解题

看了看答案，根据第三个答案，又大概优化了一下自己的代码（程序应该是可以更优的，但是现在不想改了，下次看到了，再继续优化）：

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    public ListNode deleteDuplicates(ListNode head) {            ListNode dummy=new ListNode(0);            dummy.next=head;            ListNode preNode=dummy;            ListNode curNode=head;            while(curNode!=null){                if(curNode.next!=null){                    if(curNode.next.val!=curNode.val){                        preNode=curNode;                        curNode=curNode.next;                    }else{                        curNode=curNode.next;                        while((curNode.next!=null)&&(curNode.next.val==curNode.val)){                            curNode=curNode.next;                        }                        curNode=curNode.next;                        preNode.next=curNode;                    }                 }else{                     return dummy.next;                 }                       }           return dummy.next;                 }    }

答案：

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    public ListNode deleteDuplicates(ListNode head) {        // Start typing your Java solution below        // DO NOT write main() function        if(head==null || head.next==null) return head;         ListNode dummy=new ListNode(0);         dummy.next=head;        boolean det=false;         ListNode pre=dummy;         ListNode wait=head;         ListNode runner=head.next;         while(runner!=null)        {            if(runner.val==wait.val)            {                det=true;                 wait.next=runner.next;             }else            {                if(det)                {                    pre.next=runner;                     wait=runner;                    det=false;                 }else                {                    pre=wait;                     wait=runner;                }            }            runner=runner.next;         }        if(det) pre.next=null;         return dummy.next;     }}/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    public ListNode deleteDuplicates(ListNode head) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.        ListNode dumb = new ListNode(Integer.MIN_VALUE);        dumb.next = head;        ListNode curNode = dumb;        boolean hasDup = false;        while(curNode != null && curNode.next != null && curNode.next.next != null) {        if(curNode.next.val == curNode.next.next.val) {        curNode.next = curNode.next.next;        hasDup = true;        } else {        if(!hasDup) {        curNode = curNode.next;        } else {        curNode.next = curNode.next.next;        }        hasDup = false;        }        }        if(hasDup) curNode.next = null;        return dumb.next;    }}public class Solution {    public ListNode deleteDuplicates(ListNode head) {        ListNode dummy = new ListNode(0);        dummy.next = head;        ListNode cur = dummy;        while(cur.next != null && cur.next.next != null) {            if(cur.next.val == cur.next.next.val) {                ListNode runner = cur.next.next.next;                while(runner != null && runner.val == cur.next.val) {                    runner = runner.next;                }                cur.next = runner;            } else {                cur = cur.next;            }        }        return dummy.next;    }}

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