Jelly, 8 bytes

_ƝṠÞ+SƊƑ

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How it works

_ƝṠÞ+SƊƑ  Main link. Argument: n (integer)



_Ɲ        Take the differences of neighboring digits.

          This maps n = abcd to [a-b, b-c, c-d].

       Ƒ  Fixed; apply the link to the left and return 1 if the result is equal to

          its argument, 0 if not.

      Ɗ       Drei; combine the three links to the left into a monadic chain.

  ṠÞ              Sort the differences by their signs (negative, zero, positive).

     S            Take the sum of the differences, yielding 0 if and only if the

                  first digit is equal to the last.

    +             Add the sum to each difference.

JavaScript (ES6), 62 54 bytes

Takes input as a string. Returns a Boolean value.

s=>s[-[...s].some(p=q=n=>q>(q=Math.sign(p-(p=n))))]==p

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Commented

s =>                  // s = input string

  s[                  // we will eventually access either s[0] or s[-1]

    -[...s].some(     // depending on the result of this some()

      p = q =         // initialize p and q to non-numeric values

      n =>            // for each digit n:

        q > (         //   compare q with

          q =         //   the new value of q,

          Math.sign(  //   defined as the sign of

          p - (p = n) //   the difference between the current digit and the previous one

        ))            //   yield true if the previous q is greater than the new q

    )                 // s[-1] being undefined, a truhty some() will force the test to fail

  ] == p              // otherwise: test if the 1st digit s[0] is equal to the last digit p

JavaScript (ES6), 65 bytes

A solution using a regular expression. Takes input as a string. Returns $0$ or $1$.

s=>/N(,-d+)*(,0)*[^0-]*$/.test([...s].map(p=v=>p-(p=v)))&p==s[0]

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How?

We first convert the number to a list of pairwise digit differences in $[-9,9]$:

[...s].map(p = v => p - (p = v))

Example:

"234567992" --> [ NaN, -1, -1, -1, -1, -1, -2, 0, 7 ]

This array is coerced to a string, which gives:

"NaN,-1,-1,-1,-1,-1,-2,0,7"

We apply the following regular expression:

 +-----------------------> the second 'N' of 'NaN'

 |    +------------------> a sequence of negative numbers

 |    |     +------------> a sequence of zeros

 |    |     |     +------> a sequence of positive numbers

 |    |     |     |  +---> end of string

 |    |     |     |  |

 |/¨¨¨¨¨¨/¨¨¨/¨¨¨¨|

/N(,-d+)*(,0)*[^0-]*$/

Finally, we also test if the last digit p is equal to the first digit s[0].

Pyth, 16 bytes

&SI_._MJ.+jQT!sJ

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          jQT          input in base 10

       J.+             J = differences: [3,1,4,1] -> [-2,3,-3]

    ._M                Signs of each element of J

   _                   Reverse the list

 SI                    and check if it is Invariant under Sorting.

                       If this is true, J consists of some positive numbers,

                         followed by some 0s, followed by some negative numbers,

                         which is what we want.

            !sJ        Now we check the other hill condition by ensuring

                         sum(differences) = 0; i.e. the first and last digit are equal.

&                      We take the logical AND of both conditions.

Jelly, 11 bytes

DIµṠNṢƑaS¬$

Explanation:

D               Convert to a list of Digits.

 I              Increments; compute differences between successive elements.

  µ             Start new µonadic link.

   Ṡ              Find Ṡign of each increment

    N             then negate;

     ṢƑ           is the result invariant under Ṣorting?

                  If so, the increments consist of some positive numbers,

                     followed by some 0s, followed by some negative numbers,

                     which is what we want.

       a          Logical AND this result with

        S¬$       logical NOT of the Sum of the increments.

                  If the sum of the increments is zero, first and last digits are equal.

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Perl 6, 39 bytes

{.[0]==.tail&&[<=] $_ Z<=>.skip}o*.comb

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Explanation

{ ... }o.comb  # Split into digits and feed into block

.[0]==.tail    # First element equals last

&&             # and

     $_ Z<=>.skip  # Pairwise application of three-way comparator

[<=]           # Results never decrease

Python 2, 114 112 bytes

lambda n:all((n[0]==n[-1])*sorted(set(x))==list(x)[::d]for x,d in zip(n.split(max(n)*n.count(max(n)),1),[1,-1]))

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R, 65 bytes

Takes strings. Took the idea for checking sort invariance from the Pyth answer.

function(a)!sum(d<-diff(utf8ToInt(a)))&all(sort(k<-sign(d),T)==k)

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05AB1E, 19 17 13 12 bytes

¥D.±Â{RQsO_*

-5 bytes by creating a port of @lirtosiast's Pyth answer.

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Explanation:

¥           # Push the deltas of the digits of the (implicit) input

            #  i.e. 4588774 → [1,3,0,-1,0,-3]

 D          # Duplicate this list

  .±        # Get the sign of each

            #  [1,3,0,-1,0,-3] → [1,1,0,-1,0,-1]

    Â       # Bifurcate (short for DR: Duplicate and Reverse copy)

            #  i.e. [1,1,0,-1,0,-1] → [-1,0,-1,0,1,1]

     {      # Sort the copy

            #  i.e. [-1,0,-1,0,1,1] → [-1,-1,0,0,1,1]

      R     # Reverse it

            #  i.e. [1,1,0,0,-1,-1]

       Q    # And check if they are equal

            #  i.e. [1,1,0,-1,0,-1] and [1,1,0,0,-1,-1] → 0 (falsey)

s           # Swap to get the list of deltas again

 O          # Take the sum

            #  i.e. [1,3,0,-1,0,-3] → 0

  _         # And check if it's exactly 0

            #  0 → 1 (truthy)

*           # Check if both are truthy (and output implicitly)

            #  i.e. 0 and 1 → 0 (falsey)

Â{RQ can alternatively be (Â{Q for the same byte-count, where ( negates each sign: Try it online.

J, 23 bytes

[:((0=+/)**-:*/:*)2-/]

Idea stolen from the Jelly answers. Just wanted to see how short I could make it in J.

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MATL, 12 bytes

dZSd1<AGds~*

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Explanation

Input is a string of digits. Output is a 1 or 0. The number 222222 is a hill number according to this program. Saved 2 bytes by copying Dennis' method for checking equality of the first and last digits.

d               % Takes the difference between digits

 ZS             % Calculate the sign. 

   d            % Take the difference again. 

    1<          % A number is a hill number if these differences are < 1.

      A         % Truthy iff above is all true OR if array is empty (necessary for short inputs)

       Gds      % Push the input, and sum all the differences.

          ~     % Negate

           *    % Multiply the two tests (=logical AND).

Japt, 11 bytes

Takes input as a digit array.

ä-

eUñg)«Ux

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Python 2, 53 bytes

def f(s):x=map(cmp,s,s[1:]);s[:sorted(x)==x]!=s[-1]>_

Takes input as a string. Output is via presence or absence of an exception.

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Python 2, 62 bytes

lambda s:s[:eval('<='.join(map(str,map(cmp,s,s[1:]))))]==s[-1]

Takes input as a string and returns a Boolean.

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Mathematica/Wolfram Language, 69 64 bytes

Pure function. Takes input as an integer, returns True or False.

Sort[x=Sign@-Differences[y=IntegerDigits@#]]==x&&y[[1]]==Last@y&

Explanation:

The first clause checks the "hilliness":

• 
  IntegerDigits: Get digits from integer. Store in y.


• 
  -Differences: Take successive differences and flip signs.


• 
  Sign: Replace each entry with +1 if positive, 0 if zero, and -1 if negative. Store in x.


• 
  Sort: Sort list of +1, 0, -1 from smallest to largest. Compare to original list in x.

The second clause checks whether the first and last digits are equal.

A tip of the hat to @IanMiller for tips on refining this code.

Retina 0.8.2, 52 bytes

.

$*1;$&$*1,

(1+),1

,

^(1+);(,1+;)*(,;)*(1+,;)*1,$

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.

$*1;$&$*1,

Convert each digit to unary twice, separated by ;s and terminated by ,s. However, you can then think of the result as the first digit, a ;, then all the pairs of adjacent digits, the digits of each pair separated by , and the pairs separated by ;s, then another ;, then the last digit, then a final ,.

(1+),1

,

Subtract the pairs of adjacent digits. This leaves ;,; for equal digits and 1s on the greater side for unequal digits. (This could be done as part of the following regex but obviously that wouldn't be so golfy.)

^(1+);(,1+;)*(,;)*(1+,;)*1,$

Match the first digit, then any number of pairs of ascending digits, then any number of pairs of equal digits, then any number of pairs of descending digits, then match the first digit again at the very end.

Red, 181 bytes

func[n][m: last sort copy t: s: form n

parse t[opt[copy a to m(a: sort unique a)]copy b thru any m

opt[copy c to end(c: sort/reverse unique c)]](s = rejoin[a b c])and(s/1 = last s)]

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More readable:

f: func[n][

    t: s: form n                                    

    m: last sort copy t                             

    parse t [ opt [ copy a to m (a: sort unique a) ] 

              copy b thru any m

              opt [ copy c to end (c: sort/reverse unique c) ]

            ]

    (s = rejoin [ a b c ]) and (s/1 = last s)

]

Powershell, 77 bytes

($x=-join("$($args|%{"-$_;$_"})"|iex))-match'^(-d)+0*d+$'-and$x[1]-eq$x[-1]

Less golfed test script:

$f = {

                                           # $args = 1,2,3,3,3,2,1

$a=$args|%{"-$_;$_"}                       # "-1;1","-2;2","-3;3","-3;3","-3;3","-2;2","-1;1"

$d="$a"                                    # "-1;1 -2;2 -3;3 -3;3 -3;3 -2;2 -1;1"

$x=-join($d|Invoke-Expression)             # "-1-1-100111"

$x-match'^(-d)+0*d+$'-and$x[1]-eq$x[-1]  # $true or $false



}



@(

    ,($True , 1,2,3,2,1 )

    ,($True , 1,2,3,3,3,2,1 )

    ,($True , 9,9 )

    ,($True , 3 )

    ,($True , 2,3,4,5,6,7,9,9,2 )

    ,($False, 1,2,3,2 )

    ,($False, 7,7,8,8,9,6 )

    ,($False, 2,3,2,3,2 )

    ,($False, 4,5,5,6,6,5,5,4 )

    ,($False, 5,6,4,5 )

) | % {

    $expected,$a = $_

    $result = &$f @a

    "$($result-eq$expected): $result"

}

Output:

True: True

True: True

True: True

True: True

True: True

True: False

True: False

True: False

True: False

True: False

C# (Visual C# Interactive Compiler), 161 bytes

s=>{var m=s.OrderBy(c=>c).Last();return s[0]==s.Last()&Enumerable.Range(1,s.Length-1).All(i=>i>s.LastIndexOf(m)?s[i-1]>s[i]:i>s.IndexOf(m)?m==s[i]:s[i-1]<s[i]);}

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Here is an overview of how this works...

1. Input is in the form of a string
   


2. Find the largest digit


3. Ensure the first and last digits are the same


4. Ensure digits after the last occurrence of the largest digit are decreasing


5. Ensure digits between the first and last occurrence of the largest digit are equal to the largest digit


6. Ensure digits before the first occurrence of the largest digit are increasing

Python 3, 114 bytes

def f(r):

 l=[*r]

 for i in-1,0:

  while 1<len(l)and l[i]<l[(1,-2)[i]]:l.pop(i)

 return 2>len({*l})and r[0]==r[-1]

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Way longer than some Python 2 solutions, but this one is def-based and I like it.

Ruby, 47 bytes

->n{(r=n.each_cons(2).map{|a,b|a<=>b})==r.sort}

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Input as array of digits, output is boolean.