Searching for the index of an item in a recursive linked list in java
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
add a comment |
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
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
add a comment |
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
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
java recursion linked-list
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
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);
add a comment |
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.
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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);
add a comment |
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);
add a comment |
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);
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);
edited Nov 13 '18 at 1:31
answered Nov 13 '18 at 1:25
muradmmuradm
864519
864519
add a comment |
add a comment |
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.
add a comment |
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.
add a comment |
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.
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.
answered Nov 13 '18 at 1:26
The Zach ManThe Zach Man
1146
1146
add a comment |
add a comment |
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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