Searching for the index of an item in a recursive linked list in java










1














My task is to write a wrapper and recursive method to search for a given item and return the index of that item within a linked list. This is the code I have, and it works for items that are in the list, but when given an item that is not in the list it just returns the index of the tail. Any idea what i'm doing wrong here?



public int searchIndex(E item) 
return searchIndex(item, head, 0);


private int searchIndex(E item, Node<E> node, int index)

if (node == null)
return -1;

else if (item.equals(node.data))
return 0;

else
return 1 + searchIndex(item, node.next, index);














share|improve this question





















  • what do you expect it to return if its not in the list? Cause : 1 + searchIndex you are adding 1 until you iterate the entire list when the element doesn't exist as well.
    – nullpointer
    Nov 13 '18 at 1:24
















1














My task is to write a wrapper and recursive method to search for a given item and return the index of that item within a linked list. This is the code I have, and it works for items that are in the list, but when given an item that is not in the list it just returns the index of the tail. Any idea what i'm doing wrong here?



public int searchIndex(E item) 
return searchIndex(item, head, 0);


private int searchIndex(E item, Node<E> node, int index)

if (node == null)
return -1;

else if (item.equals(node.data))
return 0;

else
return 1 + searchIndex(item, node.next, index);














share|improve this question





















  • what do you expect it to return if its not in the list? Cause : 1 + searchIndex you are adding 1 until you iterate the entire list when the element doesn't exist as well.
    – nullpointer
    Nov 13 '18 at 1:24














1












1








1







My task is to write a wrapper and recursive method to search for a given item and return the index of that item within a linked list. This is the code I have, and it works for items that are in the list, but when given an item that is not in the list it just returns the index of the tail. Any idea what i'm doing wrong here?



public int searchIndex(E item) 
return searchIndex(item, head, 0);


private int searchIndex(E item, Node<E> node, int index)

if (node == null)
return -1;

else if (item.equals(node.data))
return 0;

else
return 1 + searchIndex(item, node.next, index);














share|improve this question













My task is to write a wrapper and recursive method to search for a given item and return the index of that item within a linked list. This is the code I have, and it works for items that are in the list, but when given an item that is not in the list it just returns the index of the tail. Any idea what i'm doing wrong here?



public int searchIndex(E item) 
return searchIndex(item, head, 0);


private int searchIndex(E item, Node<E> node, int index)

if (node == null)
return -1;

else if (item.equals(node.data))
return 0;

else
return 1 + searchIndex(item, node.next, index);











java recursion linked-list






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share|improve this question











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share|improve this question










asked Nov 13 '18 at 1:19









Michael AlbertMichael Albert

103




103











  • what do you expect it to return if its not in the list? Cause : 1 + searchIndex you are adding 1 until you iterate the entire list when the element doesn't exist as well.
    – nullpointer
    Nov 13 '18 at 1:24

















  • what do you expect it to return if its not in the list? Cause : 1 + searchIndex you are adding 1 until you iterate the entire list when the element doesn't exist as well.
    – nullpointer
    Nov 13 '18 at 1:24
















what do you expect it to return if its not in the list? Cause : 1 + searchIndex you are adding 1 until you iterate the entire list when the element doesn't exist as well.
– nullpointer
Nov 13 '18 at 1:24





what do you expect it to return if its not in the list? Cause : 1 + searchIndex you are adding 1 until you iterate the entire list when the element doesn't exist as well.
– nullpointer
Nov 13 '18 at 1:24













2 Answers
2






active

oldest

votes


















3














Your conditions are wrong. Let's break down:



private int searchIndex(E item, Node<E> node, int index) 
if (node == null)
// ok you say that if list ended found index is -1, i.e. not found
return -1;

else if (item.equals(node.data))
// item is found, but your result is 0?
// your result should be index
return 0;

else
// your return is an index, so you can't make math on result
// also passing same index over and over again
return 1 + searchIndex(item, node.next, index);




For recursion, you have to state proper conditions. And normally your return is result, and arguments variations.



private int searchIndex(E item, Node<E> node, int index) 
// break condition: not found at all
if (node == null) return -1;
// break condition: found at index
if (item.equals(node.data)) return index;
// continue condition: proceed to next node and index
return searchIndex(item, node.next, index + 1);






share|improve this answer






























    0














    When you return -1, your recursive statement adds one to it and makes it indistinguishable from if the tail node had returned zero because of a match.






    share|improve this answer




















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3














      Your conditions are wrong. Let's break down:



      private int searchIndex(E item, Node<E> node, int index) 
      if (node == null)
      // ok you say that if list ended found index is -1, i.e. not found
      return -1;

      else if (item.equals(node.data))
      // item is found, but your result is 0?
      // your result should be index
      return 0;

      else
      // your return is an index, so you can't make math on result
      // also passing same index over and over again
      return 1 + searchIndex(item, node.next, index);




      For recursion, you have to state proper conditions. And normally your return is result, and arguments variations.



      private int searchIndex(E item, Node<E> node, int index) 
      // break condition: not found at all
      if (node == null) return -1;
      // break condition: found at index
      if (item.equals(node.data)) return index;
      // continue condition: proceed to next node and index
      return searchIndex(item, node.next, index + 1);






      share|improve this answer



























        3














        Your conditions are wrong. Let's break down:



        private int searchIndex(E item, Node<E> node, int index) 
        if (node == null)
        // ok you say that if list ended found index is -1, i.e. not found
        return -1;

        else if (item.equals(node.data))
        // item is found, but your result is 0?
        // your result should be index
        return 0;

        else
        // your return is an index, so you can't make math on result
        // also passing same index over and over again
        return 1 + searchIndex(item, node.next, index);




        For recursion, you have to state proper conditions. And normally your return is result, and arguments variations.



        private int searchIndex(E item, Node<E> node, int index) 
        // break condition: not found at all
        if (node == null) return -1;
        // break condition: found at index
        if (item.equals(node.data)) return index;
        // continue condition: proceed to next node and index
        return searchIndex(item, node.next, index + 1);






        share|improve this answer

























          3












          3








          3






          Your conditions are wrong. Let's break down:



          private int searchIndex(E item, Node<E> node, int index) 
          if (node == null)
          // ok you say that if list ended found index is -1, i.e. not found
          return -1;

          else if (item.equals(node.data))
          // item is found, but your result is 0?
          // your result should be index
          return 0;

          else
          // your return is an index, so you can't make math on result
          // also passing same index over and over again
          return 1 + searchIndex(item, node.next, index);




          For recursion, you have to state proper conditions. And normally your return is result, and arguments variations.



          private int searchIndex(E item, Node<E> node, int index) 
          // break condition: not found at all
          if (node == null) return -1;
          // break condition: found at index
          if (item.equals(node.data)) return index;
          // continue condition: proceed to next node and index
          return searchIndex(item, node.next, index + 1);






          share|improve this answer














          Your conditions are wrong. Let's break down:



          private int searchIndex(E item, Node<E> node, int index) 
          if (node == null)
          // ok you say that if list ended found index is -1, i.e. not found
          return -1;

          else if (item.equals(node.data))
          // item is found, but your result is 0?
          // your result should be index
          return 0;

          else
          // your return is an index, so you can't make math on result
          // also passing same index over and over again
          return 1 + searchIndex(item, node.next, index);




          For recursion, you have to state proper conditions. And normally your return is result, and arguments variations.



          private int searchIndex(E item, Node<E> node, int index) 
          // break condition: not found at all
          if (node == null) return -1;
          // break condition: found at index
          if (item.equals(node.data)) return index;
          // continue condition: proceed to next node and index
          return searchIndex(item, node.next, index + 1);







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Nov 13 '18 at 1:31

























          answered Nov 13 '18 at 1:25









          muradmmuradm

          864519




          864519























              0














              When you return -1, your recursive statement adds one to it and makes it indistinguishable from if the tail node had returned zero because of a match.






              share|improve this answer

























                0














                When you return -1, your recursive statement adds one to it and makes it indistinguishable from if the tail node had returned zero because of a match.






                share|improve this answer























                  0












                  0








                  0






                  When you return -1, your recursive statement adds one to it and makes it indistinguishable from if the tail node had returned zero because of a match.






                  share|improve this answer












                  When you return -1, your recursive statement adds one to it and makes it indistinguishable from if the tail node had returned zero because of a match.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 13 '18 at 1:26









                  The Zach ManThe Zach Man

                  1146




                  1146



























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