刚刚开始学 ruby，许多时候用什么方法还要先去查的情况～
各位走过路过、或者也在做题的童鞋欢迎前来指教，多谢多谢

# 4. Median of Two Sorted Arrays
# There are two sorted arrays nums1 and nums2 of size m and n respectively.
# Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
# You may assume nums1 and nums2 cannot be both empty.
# Example 1:
# nums1 = [1, 3]
# nums2 = [2]
# The median is 2.0
# Example 2:
# nums1 = [1, 2]
# nums2 = [3, 4]
# The median is (2 + 3)/2 = 2.5
# @param {Integer[]} nums1
# @param {Integer[]} nums2
# @return {Float}
def find_median_sorted_arrays(nums1, nums2)
  return 0 if nums1.empty? && nums2.empty?
  if nums1.empty?||nums2.empty?
    len = nums1.size | nums2.size
    if len % 2 == 1
      return nums1.empty? ? nums2[len/2] : nums1[len/2]
    elsif len % 2 == 0
      return nums1.empty? ? (nums2[len/2 - 1] + nums2[len/2]).to_f/2 : (nums1[len/2 - 1] + nums1[len/2]).to_f/2
    end
  end
  nums2.each do |i|
    nums1[nums1.size,0]=i
    # print nums1
  end
  nums11 = nums1.sort
  l = nums11.size
  return l % 2 == 0 ? (nums11[l / 2] + nums11[l / 2 - 1]).to_f / 2 : nums11[l / 2].to_f
end

# 6. ZigZag Conversion

# Input: s = "PAYPALISHIRING", numRows = 4
# Output: "PINALSIGYAHRPI"
# Explanation:

# P     I    N
# A   L S  I G
# Y A   H R
# P     I

# @param {String} s
# @param {Integer} num_rows
# @return {String}
def convert(s, num_rows)
  len = s.size
  res = ""
  # 特殊情况，如　"" , 2
  return res if len == 0
  # 特殊情况，如　"A",2 或者 "ABC" , 1 
  return s if num_rows >= len || num_rows == 1
  # addition为上例中Ｐ和Ｉ，Ｉ和Ｎ在原字符串中位置的差
  addition = 2 * num_rows - 2
  # repeat为能"写"出的Ｚ字型的个数
  repeat = (len - num_rows) / addition + 1 
  num_rows.times do |i|
    # 每次循环s[i]分别为
    # Ｐ
    # Ａ
    # Ｙ
    # Ｐ
    res = res + s[i]
    # i表示第几行，ｊ表示第几个Ｚ字型
    repeat.times do |j|
      if i==0 || i==num_rows - 1
        res = res + s[addition*(j+1)+i] if addition*(j+1)+i < len
      else
        res = res + s[addition*(j+1)-i] if addition*(j+1)-i < len
        res = res + s[addition*(j+1)+i] if addition*(j+1)+i < len 
      end
    end
  end
  return res
end

# 看到一个好理解又简洁的版本，马来学习
# position表示第几行，direction表示上一行还是下一行，在两个判断条件处决定
# 这个代码是按字符串中字符顺序分别存到buf的第几个单元中，每个单元表示一行，最后join一下
# 相比之下，我的代码是直接将结果存在res中，然后输出res
def convert(s, num_rows)
  return s if num_rows ==1
  buf = num_rows.times.collect{[]}
  position = 0
  direction=1
  s.each_char do |c|
      buf[position]<< c
      if position+direction==num_rows
          direction=-1
      elsif position+direction==-1
          direction=1
      end
      position+=direction
  end
  buf.join
end

# 8. String to Integer (atoi)
# 发现可以直接用to_i，，， 简直了
# @param {String} str
# @return {Integer}
def my_atoi(str)
  # return 0 if str.empty? || str[0] != '-' || str[0] != '+' || !((48..57) === str[0].ord)
  res = /[-|+]?[0-9]*/.match(str.lstrip).to_s.to_i
  res = res > (2**31 - 1) ? (2**31 - 1) : (res <(-2**31) ? (-2**31) : res)
  return res
end

To be continued…