LinkList 专题之三

Rotate List

Given a list, rotate the list to the right by k places, where k is non-negative.

For example:
Given 1->2->3->4->5->NULL and k = 2,
return 4->5->1->2->3->NULL.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode rotateRight(ListNode head, int k) {
        ListNode first=head;
        ListNode slow=head;
         ListNode pre=head;
        if(head==null||head.next==null||k==0){
            return head;
        }
        int count=1;
        while(pre.next!=null){
            pre=pre.next;
            count++;
        }
       
        if(k==count||k%count==0){ //注意k>list AND k为list的倍数
            return head;
        }
        k=k%count;
        for(int i=0;i<k;i++){
            first=first.next;
        }
        while(first!=null){
            slow=slow.next;
            first=first.next;
        }
        ListNode res=slow;
        while(slow.next!=null){
            slow=slow.next;
        }
        ListNode temp=head;
        while(temp!=res){
            slow.next=temp;
            temp=temp.next;
            slow=slow.next;
        }
        slow.next=null;
        return res;
    }
}

time O（n）

Remove Duplicates from Sorted List

Given a sorted linked list, delete all duplicates such that each element appear only once.

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

分析：两个指针遍历pre和current，一前一后

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
         if(head==null||head.next==null){
            return head;
        }
       ListNode newhead=new ListNode(0);
       ListNode res=newhead;
       ListNode pre=head;
       ListNode current=head.next;
       while(current!=null){
           if(pre.val!=current.val){
               newhead.next=new ListNode(pre.val);
               pre=pre.next;
               current=current.next;
               newhead=newhead.next;
           }else{
               current=current.next;
               pre=pre.next;
           }
       }
          newhead.next=pre;
          return res.next;
    }
}

Remove Duplicates from Sorted List

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.

分析：加一个可识别的flag去掉重复元素以及一个特例［1，1］or［2，2］....返回为null

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if(head==null||head.next==null){
            return head;
        }
       ListNode newhead=new ListNode(0);
       ListNode res=newhead;
       ListNode pre=head;
       ListNode current=head.next;
       if(pre.val==current.val&&current.next==null){
           return null;
       }
       boolean flag=true;
       while(current!=null){
           if(pre.val!=current.val&& flag==true){
               newhead.next=new ListNode(pre.val);
               pre=pre.next;
               current=current.next;
               newhead=newhead.next;
           }else if(pre.val!=current.val&& flag==false){
               current=current.next;
               pre=pre.next;
               flag=true;
           }else if(pre.val==current.val){
               current=current.next;
               pre=pre.next;
               flag=false;
           }
       }
       if(flag==true){
           newhead.next=pre;
       }
       return res.next;
    }
}

Partition List

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

You should preserve the original relative order of the nodes in each of the two partitions.

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

分成两个链表

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode partition(ListNode head, int x) {
        ListNode small=new ListNode(0);
        ListNode big=new ListNode(0);
        ListNode b=big;
        ListNode res= small;
        if(head==null||head.next==null){
            return head;
        }
        ListNode temp=head;
        while(temp!=null){
            if(temp.val<x){
              small.next=temp;
              small=small.next;
              temp=temp.next;
            }else{
             big.next= temp;
             big=big.next;
             temp=temp.next;
            }
        }
        big.next=null;
        if(b.next!=null){
            small.next=b.next;
        }
        return res.next;
    }
}