Solution 1: Walk the linked list to find the length of the list. Lets say n is the length of the list. Walk the list again upto ⌊ n/2 ⌋.

Solution 2: Use two pointers. Move one pointer at twice the speed of the second. When the 1st pointer reaches the end of the list, the 2nd pointer will be pointing to the middle node. Note: If the list has even number of nodes, the middle node will be floor of ⌊ n/2 ⌋.

Code:

Node * FindMiddle(Node *listHead) { Node *ptr1, *ptr2; // we need 2 pointers ptr1 = ptr2 = listHead; // set the pointers to point to the list head initially int i=0; while(ptr1->next != NULL) // keep looping until we reach the tail // (next will be NULL for the last node) { if(i == 0) { ptr1 = ptr1->next; //increment only the 1st pointer i=1; } else if( i == 1) { ptr1 = ptr1->next; //increment both pointers ptr2 = ptr2->next; i = 0; } } return ptr2; //now return the ptr2 which points to the middle node }

My logic:

ReplyDeleteptr1 = ptr2 = head;

tmp = NULL;

while( ptr1 && (tmp=ptr2->next) && (ptr2=tmp->next) ptr1=ptr1->next;

return ptr1;