每天一个算法之链表逆序

需要4个指针，当前节点，前一个节点，后一个节点和尾节点（新头），当前节点不断后移，不断把next执行前一个

class listnode{

int data;

listnode next;

}

public class reverse {

public static void main(String args[]){

int[] n={1,3,5,7,9,11,13,15,17};

//int[] a={1,2,3,4,5,6,7,8,9};

//reorder(a);

//for(int each:a){

// System.out.print(each+" ");

//}

listnode list=CreateList(n);

listnode revlist=reverse(list);

print(revlist);

}

public static listnode CreateList(int[] n){

listnode head=new listnode();

head.data=n[0];

head.next=null;

listnode point=head;

for(int i=1;i<n.length;i++){

listnode node=new listnode();

node.data=n[i];

node.next=null;

point.next=node;

point=point.next;

}

return head;

}

public static void print(listnode l){

while(l!=null){

System.out.print(l.data+" ");

l=l.next;

}

}

public static listnode reverse(listnode list){

listnode tail=null;

listnode node=list;

listnode pre=null;

while(node!=null){

listnode next=node.next;

if(next==null)

tail=node;

node.next=pre;

pre=node;

node=next;

}

return tail;

}

}