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    <p>The key to this algorithm is to set two pointers <code>p1</code> and <code>p2</code> apart by <code>n-1</code> nodes initially so we want <code>p2</code> to point to the <code>(n-1)th</code> node from the start of the list then we move <code>p2</code> till it reaches the <code>last</code> node of the list. Once <code>p2</code> reaches end of the list <code>p1</code> will be pointing to the nth node from the end of the list.</p> <p>I've put the explanation inline as comments. Hope it helps:</p> <pre><code>// Function to return the nth node from the end of a linked list. // Takes the head pointer to the list and n as input // Returns the nth node from the end if one exists else returns NULL. LinkedListNode nthToLast(LinkedListNode head, int n) { // If list does not exist or if there are no elements in the list,return NULL if (head == null || n &lt; 1) { return null; } // make pointers p1 and p2 point to the start of the list. LinkedListNode p1 = head; LinkedListNode p2 = head; // The key to this algorithm is to set p1 and p2 apart by n-1 nodes initially // so we want p2 to point to the (n-1)th node from the start of the list // then we move p2 till it reaches the last node of the list. // Once p2 reaches end of the list p1 will be pointing to the nth node // from the end of the list. // loop to move p2. for (int j = 0; j &lt; n - 1; ++j) { // while moving p2 check if it becomes NULL, that is if it reaches the end // of the list. That would mean the list has less than n nodes, so its not // possible to find nth from last, so return NULL. if (p2 == null) { return null; } // move p2 forward. p2 = p2.next; } // at this point p2 is (n-1) nodes ahead of p1. Now keep moving both forward // till p2 reaches the last node in the list. while (p2.next != null) { p1 = p1.next; p2 = p2.next; } // at this point p2 has reached the last node in the list and p1 will be // pointing to the nth node from the last..so return it. return p1; } </code></pre> <p>Alternatively we can set <code>p1</code> and <code>p2</code> apart by n nodes instead of <code>(n-1)</code> and then move <code>p2</code> till the end of the list instead of moving till the last node:</p> <pre><code>LinkedListNode p1 = head; LinkedListNode p2 = head; for (int j = 0; j &lt; n ; ++j) { // make then n nodes apart. if (p2 == null) { return null; } p2 = p2.next; } while (p2 != null) { // move till p2 goes past the end of the list. p1 = p1.next; p2 = p2.next; } return p1; </code></pre>
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    1. POHow to find nth element from the end of a singly linked list?
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    1. COThanks for neat explanation
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    2. COnice explanation... thanks
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