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Diffstat (limited to 'leetcode/lc-9-palindromic-number.rkt')
| -rw-r--r-- | leetcode/lc-9-palindromic-number.rkt | 38 |
1 files changed, 0 insertions, 38 deletions
diff --git a/leetcode/lc-9-palindromic-number.rkt b/leetcode/lc-9-palindromic-number.rkt deleted file mode 100644 index 1fccbef..0000000 --- a/leetcode/lc-9-palindromic-number.rkt +++ /dev/null @@ -1,38 +0,0 @@ -#lang racket -(define/contract (is-palindrome x) - (-> exact-integer? boolean?) - ; get the easy cases out of the way first - ; negative numbers are not palindromes, single-digit numbers are - (cond [(x . < . 0) #false] - [(x . < . 10) #true] - [else - ; order-of-magnitude returns the scientific notation exponent - ; so add 1 to get the number of digits - (define digits - (add1 (order-of-magnitude x))) - ; figure out how many digits we need to trim to find the mirrored halves - ; if there are an even number of digits 2n, we will remove n of them from the right - ; if there are an odd number of digits 2n+1, we will remove n+1 of them from the right - (define half-digits - (cond [(even? digits) (/ digits 2)] - [else (/ (add1 digits) 2)])) - ; divide x by a power of 10 to get the digits to match - (define front-half - (quotient x (expt 10 half-digits))) - ; reverse the back half with repeated divisions by 10 - (define back-half-reversed - (for/fold ([reversed 0] - [remaining (remainder x (expt 10 half-digits))] - ; build up the sum of the digits in reversed and return it at the end - #:result reversed) - ; if we have an odd number of digits, we don't need to match the middle one - ([n (in-range (if (even? digits) - half-digits - (sub1 half-digits)))]) - ; shift all the accumulated digits in the mirror to the left one place - ; and add the next one to the right, - ; then chop the right-most digit off the original - (values (+ (* 10 reversed) (remainder remaining 10)) - (quotient remaining 10)))) - ; finally, check to see if the mirrored right is equal to the original left - (= front-half back-half-reversed)])) |