We love our weird puzzles, us Brits

Question

In a few British newspapers there is a game known as Hidato. It is somewhat similar to Sudoku, albeit instead of having 1-9 in a line and block, it's about placing numbers such that they connect in order from 01 all the way up to the highest one, so they are all touching horizontally, diagonally or vertically.

Inputs will contain multiple lines separated by \n, containing blocks separated by a space, which you can assume to be two characters wide. Each block will be a number, a blank space to be filled (indicated by --), or a wall that cannot have numbers in (XX).

Your output should match the provided one albeit with empty blocks provided with numbers. Note that there may not be a unique, or even the existence of, a solution – some may yield multiple due to their ambiguity, much like Sudoku, and some may be literally unsolvable, in which case you should give a falsey output, but you can assume inputs are formatted as below.

Use a standard header Language: XX bytes. Happy golfing!

Examples

Inputs 01 XX 03, 01 -- 04, 01 --, etc should all return something falsey.

Input:

01 -- --
-- XX 05

Output:

01 03 04
02 XX 05

Input:

-- 33 35 -- -- XX XX XX    
-- -- 24 22 -- XX XX XX      
-- -- -- 21 -- -- XX XX
-- 26 -- 13 40 11 XX XX
27 -- -- -- 09 -- 01 XX
XX XX -- -- 18 -- -- XX
XX XX XX XX -- 07 -- --
XX XX XX XX XX XX 05 --

Output:

32 33 35 36 37 XX XX XX
31 34 24 22 38 XX XX XX
30 25 23 21 12 39 XX XX
29 26 20 13 40 11 XX XX
27 28 14 19 09 10 01 XX
XX XX 15 16 18 08 02 XX
XX XX XX XX 17 07 06 03
XX XX XX XX XX XX 05 04

Input:

XX XX XX XX -- 53 XX XX XX XX
XX XX XX XX -- -- XX XX XX XX
XX XX 56 -- -- -- 30 -- XX XX
XX XX -- -- -- -- -- -- XX XX
XX -- -- 20 22 -- -- -- -- XX
XX 13 -- 23 47 -- 41 -- 34 XX
-- -- 11 18 -- -- -- 42 35 37
-- -- -- -- 05 03 01 -- -- --
XX XX XX XX -- -- XX XX XX XX
XX XX XX XX 07 -- XX XX XX XX

Output:

XX XX XX XX 52 53 XX XX XX XX
XX XX XX XX 54 51 XX XX XX XX
XX XX 56 55 28 50 30 31 XX XX
XX XX 26 27 21 29 49 32 XX XX
XX 25 24 20 22 48 45 44 33 XX
XX 13 19 23 47 46 41 43 34 XX
14 12 11 18 04 02 40 42 35 37
15 16 17 10 05 03 01 39 38 36
XX XX XX XX 09 06 XX XX XX XX
XX XX XX XX 07 08 XX XX XX XX

Answer 1

JavaScript (Node.js), 482 bytes

This is a brute-force solution, it starts at 01 and checks every neighbouring cell checking for empty cells (--) or the desired number and following the path to completion or impossibility. If the desired number exists and isn't a neighbour it shortcuts this solution. Takes a few seconds for the largest grid.

This probably isn't particularly interesting, but I thought I'd try my hand at making a solution before looking at the answers linked on Rosetta Code and I enjoyed tackling a slightly more difficult challenge!

Finds all solutions when many exist. The body is a function that accepts a two dimensional array and the footer processes the input to the desired format, and returns the result to the desired format too. Happy to provide more information (and a less golfed implementation if there's interest).

f=a=>{F=(D,n,j)=>[Z=[].concat(...D),z=Z.indexOf(j),z>-1&&[x=z%w,y=z/w|0],z>-1&&[[x-1,y-1],[x,y-1],[x+1,y-1],[x-1,y],[x+1,y],[x-1,y+1],[x,y+1],[x+1,y+1]]][n];C=q=>q.map(Q=>Q.slice());w=a[0][L='length'];l=F(a,0).filter(c=>c!='XX')[L];R=[];r=(s,d)=>{let n=`0${+s+1}`.slice(-2);N=F(d,2,n);n>l?R.push(C(d)):~F(d,1,s)?(p=F(d,3,s),p.filter(P=>P==N+'')[L]?r(n,C(d)):!~F(d,1,n)?p.map(I=>{[x,y]=I,(x<0||x>w-1||y<0||y>d[L]-1)||d[y][x]=='--'&&(D=C(d),r(D[y][x]=n,D))}):0):0};r('01',a);return R}

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Answer 2

Python3, 860 bytes:

R=range
S=lambda x,y:[(X,Y)for X in R(x)for Y in R(y)]
def O(b,x,y,X,Y):
 for A,B in[(0,1),(0,-1),(1,0),(-1,0),(1,1),(1,-1),(-1,1),(-1,-1)]:
  if-1<(j:=x+A)<X and-1<(k:=y+B)<Y and b[j][k]!='XX':yield(j,k,b[j][k])
def f(b):
 b=[[int(j)if j.isdigit()else j for j in i.split()]for i in b.split('\n')]
 k=S(*(I:=(len(b),len(b[0]))))
 if len(d:=[(x,y)for x,y in k if type(b[x][y])==int])==1:return
 s,e=min(d,key=(_:=lambda x:b[x[0]][x[1]])),max(d,key=_)
 q,a=[(s,{(x,y)for x,y in k if b[x][y]=='--'},b)],[]
 D=[b[x][y]for x,y in d]
 while q:
  (X,Y),W,B=q.pop(0)
  if(X,Y)==e and W==set():return'\n'.join(' '.join(str(j).zfill(2)for j in i)for i in B)
  T=[]
  for x,y,C in O(B,X,Y,*I):
   if C==B[X][Y]+1:q+=[((x,y),W,B)];T=[];break
   if'--'==C and B[X][Y]+1 not in D:
    U=eval(str(B));U[x][y]=B[X][Y]+1
    if U not in a:T+=[((x,y),W-{(x,y)},U)];a+=[U]
  q+=T

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