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[英]Reversing of a number using recursion gives wrong result for numbers with 3 digits or more
[英]reversing the linkedlist using recursion generating wrong output
我的反向鏈接列表的遞歸方法有什么問題嗎? 因為我得到以下輸出,反轉后僅輸出1:
原始鏈接列表: 1-> 2-> 3-> 4-> 5-> Tail
使用遞歸反向鏈接列表: 1-> Tail
public class ReverseList {
public static List ReverseRecursion(List head){
List current = head;
if(current == null){
return null;
}
if(current.next == null){
head = current;
return head;
}
ReverseRecursion(current.next);
current.next.next = current;
current.next = null;
return head;
}
public static void main (String[] args){
// Created a Single LinkedList
List myList = new List(1);
myList.next = new List(2);
myList.next.next = new List(3);
myList.next.next.next = new List(4);
myList.next.next.next.next = new List(5);
System.out.println("Original LinkedList: \n"+myList.toString());
System.out.println("Reversed LinkedList Using Recursion: \n"+ReverseRecursion(myList));
}
}
class List {
int value;
List next;
public List(int k){
value = k;
next = null;
}
public String toString(){
List cur = this;
String output = "";
while(cur != null){
output+=cur.value+"-->";
cur = cur.next;
}
return output+"Tail";
}
}
在ReverseRecursion
,你永遠不分配反轉的列表回到head
。 更改此行:
ReverseRecursion(current.next);
對此:
head = ReverseRecursion(current.next);
您離工作代碼不太遠:
public static List ReverseRecursion(List head){
List newHead;
if(head == null){
return null;
}
if(head.next == null){
return head;
}
newHead = ReverseRecursion(head.next);
head.next.next = head;
head.next = null;
return newHead;
}
見REPL
關鍵點:
current
, head
是不變的。
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