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POreverse half of the list
 

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  1. POreverse half of the list
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    <p>I was asked to write a function reverseHalves() that rearranges a given linked list so that the first half of the nodes is moved to the back of the second half. e.g. </p> <p>Given a linked list <code>[1 2 3 4 5 6]</code>, the resulting list should be <code>[4 5 6 1 2 3]</code>.<br> When given a linked list with an <em>odd number of nodes</em>, the list should be split with the first half having an additional node. That is, given the list <code>[1 2 3 4 5]</code>, the resulting list should be <code>[4 5 1 2 3]</code>. </p> <p>But the my function will give a infinite output...</p> <pre><code>typedef struct _listnode{ int item; struct _listnode *next; } ListNode; typedef struct _linkedlist{ int size; ListNode *head; ListNode *tail; } LinkedList; </code></pre> <p>Here are the functions I used:</p> <pre><code>// printList will print out the value in every nodes until there is a NULL void printList(LinkedList *ll); ListNode * findNode(LinkedList *ll, int index); int insertNode(LinkedList *ll, int index, int value); void reverseHalves(LinkedList *ll) { int index; ListNode *new_head, *new_tail; new_head = NULL; new_tail = NULL; // determine the index of new tail, and the new head which is index+1* index = (ll-&gt;size + 1) / 2; // get the new head by findNode func,whose index is index+1 // make new_head point to the node found* new_head = findNode(ll, index); // make initial tail-&gt;next be the initial head* ll-&gt;tail-&gt;next = ll-&gt;head; // set the head to be the new head ll-&gt;head = new_head; insertNode(ll, ll-&gt;size, -1); new_tail = findNode(ll, ll-&gt;size); new_tail = NULL; } </code></pre>
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    1. COWhat's an infinite output?
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    2. COAlso, with the whitespace usage you've presented, I'm not surprised even you don't understand your own code...
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    3. COwhen I tried to print the list, it will keep printing. But I have set the tail to be NULL...
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      1. USBaller
 

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