Linked List专题一(cc charpter2)

Linklist

定义：node includes pointer(next) and value

区别（与Array）:Array has indice,which is in sequence,LinkList has pointer to next.

single list and double list. 



package Linkedlist;




 




public class ListNode {




ListNode next=null;




   int val;




 




public ListNode(int d){




   this.val=d;




   }




   public void appenttotail(int d){




   ListNode end=new ListNode(d);




   ListNode n=this;




   while(n.next!=null){




    n=n.next;




   }




   n.next=end;  




   }




   public ListNode DelectNode(ListNode head,int d){




   ListNode n=head;




   if(n.next==null){




    return head.next;




   }else{




    while(n.next!=null){




     if(n.val==d){




      n.next=n.next.next;




      return head;




     }




     n=n.next;




    }




    return head;




   }




   }




   




}


 




2.Add Two Numbers


You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.


Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8




package Linkedlist;

public class ADD2NUM {
    public ListNode addTwoNumbers(ListNode head1, ListNode head2) {
         if(head1==null){
          return head2;   
         }
          if(head2==null){
          return head1;   
         }
         int carry=0;
         ListNode newhead=new ListNode(-1);
         ListNode res=newhead;
         while(head1!=null||head2!=null){
         if(head1!=null)
         {
              carry=carry+head1.val;
              head1=head1.next;
         }
         if(head2!=null)
         {
             carry=carry+head2.val;
             head2=head2.next;
         }
             res.next=new ListNode((carry)%10);
             carry=carry/10;
             res=res.next;
         }
          if(carry==1)
         {
             res.next =new ListNode(1);
             
         }
         return newhead.next;
        }
}




19.Remove Nth Node From End of List


Given a linked list, remove the nth node from the end of list and return its head.


For example,


   Given linked list: 1->2->3->4->5, and n = 2.

   After removing the second node from the end, the linked list becomes 1->2->3->5.
 暴力版




public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
       ListNode res=head;
       ListNode fina=res;
       if(head==null||(head.next==null&&n==1)){
           return null;
       }
       if(n==0){
           return head;
       }
       int count=0;
       if(n==1){
        while(head.next.next!=null){
           head=head.next;
       }  
       head.next=null;
       return fina;
       }
       if(n!=1){
       while(head!=null){
           head=head.next;
           count++;
       }
       }
       if(n==count){
           return fina.next;
       }
       int j=1;
        while(res.next!=null){
           if(j==(count-n)){
               res.next=res.next.next;
               return fina;
           }
           res=res.next;
           j++;
       }
       return fina;
    }
}




双指针




/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
       ListNode res=head;
       ListNode  fal=head;
       if(head==null||(head.next==null&&n==1)){
           return null;
       }
       if(n==0){
           return head;
       }
       for(int i=0;i<n;i++){
           res=res.next;
       }
       if(res==null){
           return head.next;
       }
       while(res.next!=null){
          res=res.next;
          fal=fal.next;
       }
       fal.next=fal.next.next;
       return head;
    }
}




Merge Two Sorted Lists


Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists. 




/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
    ListNode newhead=new ListNode(0);
    ListNode l3=newhead;
    if(l1==null){
        return l2;
    }
    if(l2==null){
        return l1;
    }
    
    while(l1!=null&&l2!=null){
        if(l1.val<l2.val){
                l3.next=new ListNode(l1.val);
                l1=l1.next;
            }else{
                l3.next=new ListNode(l2.val);
                l2=l2.next;
        }
        l3=l3.next;
    }
     if(l1==null&&l2!=null){
          l3.next=l2;
     }
      if(l2==null&&l1!=null){
          l3.next=l1;
      }
    return newhead.next;
    }
}






Merge k Sorted Lists

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.


 


 


 

归并操作(merge)，也叫归并算法，指的是将两个已经排序的序列合并成一个序列的操作。归并排序算法依赖归并操作。


归并操作的过程如下：


    

  1. 申请空间，使其大小为两个已经排序序列之和，该空间用来存放合并后的序列
    2. 

  2. 设定两个指针，最初位置分别为两个已经排序序列的起始位置
    3. 

  3. 比较两个指针所指向的元素，选择相对小的元素放入到合并空间，并移动指针到下一位置
    4. 

  4. 重复步骤3直到某一指针到达序列尾
    5. 

  5. 将另一序列剩下的所有元素直接复制到合并序列尾
    6. 






最差時間複雜度
Theta(nlog n)



最優時間複雜度
Theta(n)



平均時間複雜度
Theta(nlog n)



最差空間複雜度
Theta(n)













public int[] mergeSort(int[] arr){
    if(arr.length<2||arr == null)
        return arr;
    
    MSort(arr,0,arr.length-1);
}

public int[] MSort(int[] arr, int low, int high){
        if(low < high){
                int mid = (low+high)/2;
                int[] left = MSort(arr,low,mid);
                int[] right = MSort(arr,mid+1,high);
                return mergeTwoList(left,right);
          }  
}


public int[] mergeTwoList(int[] A, int[] B) {
    int[] C = new int[A.length + B.length];
    int k = 0;
    int i = 0;
    int j = 0;
    while(i < A.length && j < B.length) {
        if (A[i] < B[j])
            C[k++] = A[i++];
        else
            C[k++] = B[j++];
    }
    while (i < A.length) 
        C[k++] = A[i++];
    while (j < B.length) 
        C[k++] = B[j++];
    return C;
}




Merge k Sorted Lists




/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
       ListNode res;
       if(lists.length<=0){
           return null;
       }
       if(lists.length==1){
           return lists[0];
       }
       res=mergeKlist(lists,0,lists.length-1);
       return res;
    }
    public ListNode mergeKlist(ListNode[] lists,int low,int high){
        if(low<high){
            int mid=(low+high)/2;
            ListNode left=mergeKlist(lists,low,mid);
            ListNode right=mergeKlist(lists,mid+1,high);
            return mergetwo(left,right);
        }
        return lists[low];
    }
    public ListNode mergetwo(ListNode l1,ListNode l2){
        ListNode newhead=new ListNode(0);
        ListNode l3=newhead;
        if(l1==null){
            return l2;
        }
        if(l2==null){
            return l1;
        }
        while(l1!=null&&l2!=null){
            if(l1.val>l2.val){
                l3.next=l2;
                l2=l2.next;
            }else{
                l3.next=l1;
                l1=l1.next;
            }
            l3=l3.next;
        }
        if(l1!=null){
            l3.next=l1;
        }
        if(l2!=null){
            l3.next=l2;
        }
        return newhead.next;
    }
}












        

	      

	    

	  

	  

	    
      
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          原文地址：https://www.cnblogs.com/joannacode/p/4555995.html