## Python: palindrome checking function

I created a reasonable palindrome checking function in python.

**Method 1**

def ispalindrome(num): n=str(num) while len(n)>1: print n if n[0]!=n[-1]: return 0 n=n[1:-1] return 1

I thought that it would be faster avoiding a string conversion, and to somehow use the modulus (modulo) function. However when I came to write it, I found it quite difficult to code, and I’m sure there must be a better way.

**Method 2**

def ispalindrome2(num): l=1 while num/10**l>=1.0: l+=1 r=0 d=[] for i in range(1,l+1): p=num%10**i-r r+=p p=p/10**(i-1) d.append(p) for i in range(0,l/2): if d[i]!=d[-i-1]: return 0 return 1

Doing the speed tests show that Method 1 is over 5.2 times faster than Method 2.

Method 1 0.355437994003 Seconds elapsed Method 2 1.85815691948 Seconds elapsed

**Update: Mike of mikemeat sent me his method using slices (Method 3) and I adapted it slight (Method 4), then shortly after I realised it could be even more efficient by not needing to differentiate between even and odd strings (Method 5).**

**Method 3**

def ispalindrome3(x): z = str(x) if len(z)%2 == 0 and z[:len(z)/2]==z[-len(z)/2:][::-1]: return 1 if len(z)%2 != 0 and z[:(len(z)- 1)/2]==z[(-len(z) + 1)/2:][::-1]: return 1 else: return 0

**Method 4**

def ispalindrome4(x): z = str(x) if z[:len( z)/2]==z[len( z)/2+len( z)%2:][::-1]: return 1 return 0

**Method 5**

def ispalindrome5(x): z = str(x) l=len(z)/2 if z[:l]==z[-l:][::-1]: return 1 return 0

Method 1 0.357168912888 Seconds elapsed Method 2 1.83943104744 Seconds elapsed Method 3 0.179126977921 Seconds elapsed Method 4 0.179482936859 Seconds elapsed Method 5 0.149376153946 Seconds elapsed