g0001_0100.s0007_reverse_integer.solution.rb Maven / Gradle / Ivy

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104 LeetCode algorithm problem solutions

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# #Medium #Top_Interview_Questions #Math #Udemy_Integers
# #2023_11_14_Time_58_ms_(75.27%)_Space_211.2_MB_(14.29%)

# @param {Integer} x
# @return {Integer}
def reverse(x)
  new_a = x.to_s.reverse.to_i
  if x >= 0
    new_a < (2 ** 31 - 1) ? new_a : 0
  else
    -new_a > (-2) ** 31 ? -new_a : 0
  end
end