(head)
return True
This function builds up a dictionary where the keys are the values of the list,
    and the values are the positions at which these values occur in the list.
    We then iterate over the dict and if there is more than one key with an odd
    number of occurrences, bail out and return False.
    Otherwise, we want to ensure that the positions of occurrence sum to the
    value of the length of the list - 1, working from the outside of the list inward.
    For example:
    Input: 1 -> 1 -> 2 -> 3 -> 2 -> 1 -> 1
    d = {1: [0,1,5,6], 2: [2,4], 3: [3]}
    '3' is the middle outlier, 2+4=6, 0+6=6 and 5+1=6 so we have a palindrome.
This function builds up a dictionary where the keys are the values of the list,
    and the values are the positions at which these values occur in the list.
    We then iterate over the dict and if there is more than one key with an odd
    number of occurrences, bail out and return False.
    Otherwise, we want to ensure that the positions of occurrence sum to the
    value of the length of the list - 1, working from the outside of the list inward.
    For example:
    Input: 1 -> 1 -> 2 -> 3 -> 2 -> 1 -> 1
    d = {1: [0,1,5,6], 2: [2,4], 3: [3]}
    '3' is the middle outlier, 2+4=6, 0+6=6 and 5+1=6 so we have a palindrome.
This function builds up a dictionary where the keys are the values of the list and the values are the positions at which these values occur in the list . We then iterate over the dict and if there is more than one key with an odd number of occurrences bail out and return False . Otherwise we want to ensure that the positions of occurrence sum to the value of the length of the list - 1 working from the outside of the list inward . For example : Input : 1 - > 1 - > 2 - > 3 - > 2 - > 1 - > 1 d = { 1 : [ 0 1 5 6 ] 2 : [ 2 4 ] 3 : [ 3 ] } 3 is the middle outlier 2 + 4 = 6 0 + 6 = 6 and 5 + 1 = 6 so we have a palindrome .
def is_palindrome_dict(head):
    """
    This function builds up a dictionary where the keys are the values of the list,
    and the values are the positions at which these values occur in the list.
    We then iterate over the dict and if there is more than one key with an odd
    number of occurrences, bail out and return False.
    Otherwise, we want to ensure that the positions of occurrence sum to the
    value of the length of the list - 1, working from the outside of the list inward.
    For example:
    Input: 1 -> 1 -> 2 -> 3 -> 2 -> 1 -> 1
    d = {1: [0,1,5,6], 2: [2,4], 3: [3]}
    '3' is the middle outlier, 2+4=6, 0+6=6 and 5+1=6 so we have a palindrome.
    """
    if not head or not head.next:
        return True
    d = {}
    pos = 0
    while head:
        if head.val in d.keys():
            d[head.val].append(pos)
        else:
            d[head.val] = [pos]
        head = head.next
        pos += 1
    checksum = pos - 1
    middle = 0
    for v in d.values():
        if len(v) % 2 != 0:
            middle += 1
        else:
            step = 0
            for i in range(0, len(v)):
                if v[i] + v[len(v) - 1 - step] != checksum:
                    return False
                step += 1
        if middle > 1:
            return False
    return True
def is_palindrome_dict ( head ) : if not head or not head . next : return True d = { } pos = 0 while head : if head . val in d . keys ( ) : d [ head . val ] . append ( pos ) else : d [ head . val ] = [ pos ] head = head . next pos += 1 checksum = pos - 1 middle = 0 for v in d . values ( ) : if len ( v ) % 2 != 0 : middle += 1 else : step = 0 for i in range ( 0 , len ( v ) ) : if v [ i ] + v [ len ( v ) - 1 - step ] != checksum : return False step += 1 if middle > 1 : return False return True