Merge branch 'MarkKuang1991-ruby-solutions-mkuang'

Author: yanglbme

.gitignore

@@ -0,0 +1 @@
+.idea/**/*

solution/002.Add Two Numbers/Solution.rb

@@ -10,6 +10,7 @@
 # @param {ListNode} l1
 # @param {ListNode} l2
 # @return {ListNode}
+
 def add_two_numbers(l1, l2)
   return l2 if l1 == nil
   return l1 if l2 == nil
@@ -20,32 +21,17 @@ def add_two_numbers(l1, l2)
 
   l1 = l1.next
   l2 = l2.next
-  while !l1.nil? && !l2.nil?
-    cur_val = l1.val + l2.val + add
-    tmp.next = ListNode.new(cur_val % 10)
-    tmp = tmp.next
-    add = cur_val >= 10 ? 1 : 0
-
-    l1 = l1.next
-    l2 = l2.next
-  end
-
-  until l1.nil?
-    cur_val = l1.val + add
+  while !l1.nil? || !l2.nil? || add > 0
+    cur_val = add
+    cur_val += l1.nil? ? 0 : l1.val
+    cur_val += l2.nil? ? 0 : l2.val
     tmp.next = ListNode.new(cur_val % 10)
     tmp = tmp.next
     add = cur_val >= 10 ? 1 : 0
-    l1 = l1.next
-  end
 
-  until l2.nil?
-    cur_val = l2.val + add
-    tmp.next = ListNode.new(cur_val % 10)
-    tmp = tmp.next
-    add = l2.val + add >= 10 ? 1 : 0
-    l2 = l2.next
+    l1 = l1.nil? ? l1 : l1.next
+    l2 = l2.nil? ? l2 : l2.next
   end
 
-  tmp.next = ListNode.new(1) if add == 1
   l3
 end

solution/007.Reverse Integer/Solution.rb

@@ -0,0 +1,21 @@
+# @param {Integer} x
+# @return {Integer}
+def reverse(x)
+  neg = x < 0
+
+  x = x.abs
+  s = ''
+
+  x /= 10 while x > 0 && (x % 10).zero?
+
+  while x > 0
+    s += (x % 10).to_s
+    x /= 10
+  end
+
+  s = neg ? '-' + s : s
+
+  # have to explicitly constraint the int boundary as per the dummy test case
+  res = s.to_i
+  res <= 214_748_364_7 && res >= -214_748_364_8 ? res : 0
+end

solution/013.Roman to Integer/Solution.rb

@@ -0,0 +1,32 @@
+# @param {String} s
+# @return {Integer}
+def roman_to_int(s)
+  hash = Hash[
+      'I' => 1,
+      'V' => 5,
+      'X' => 10,
+      'L' => 50,
+      'C' => 100,
+      'D' => 500,
+      'M' => 1000,
+      'IV' => 4,
+      'IX' => 9,
+      'XL' => 40,
+      'XC' => 90,
+      'CD' => 400,
+      'CM' => 900
+  ]
+  res = 0
+  i = 0
+  while i < s.length
+    if i < s.length - 1 && !hash[s[i..i+1]].nil?
+      res += hash[s[i..i+1]]
+      i += 2
+    else
+      res += hash[s[i]]
+      i += 1
+    end
+  end
+
+  res
+end

solution/014.Longest Common Prefix/Solution.rb

@@ -0,0 +1,25 @@
+# @param {String[]} strs
+# @return {String}
+def longest_common_prefix(strs)
+  return '' if strs.nil? || strs.length.zero?
+
+  return strs[0] if strs.length == 1
+
+  idx = 0
+  while idx < strs[0].length
+    cur_char = strs[0][idx]
+
+    str_idx = 1
+    while str_idx < strs.length
+      return idx > 0 ? strs[0][0..idx-1] : '' if strs[str_idx].length <= idx
+
+      return '' if strs[str_idx][idx] != cur_char && idx.zero?
+      return strs[0][0..idx - 1] if strs[str_idx][idx] != cur_char
+      str_idx += 1
+    end
+
+    idx += 1
+  end
+
+  idx > 0 ? strs[0][0..idx] : ''
+end