增加寻找最小的k个数问题解法一、二的实现，修订解法三的实现

Author: Jiang

ebook/code/python/2.1：寻找最小的k个数.py

@@ -1,24 +1,70 @@
-import heapq
+# -*- coding: utf-8 -*-
 
+# bonus for Python
+# from heapq import nsmallest
+# nsmallest(n, iterable, key=None)
 
-def wrap(x):
-    return -x
+# all example are compatible with both Python 2.7.X and Python 3.X
+from heapq import heappush, heappushpop
 
+# 等价于ksmallest1 = lambda lst, k : sorted(lst)[:k]
+def ksmallest1(lst, k):
+  """
+  code example for section 2.1 solution 1
+  see https://github.com/julycoding/The-Art-Of-Programming-By-July/blob/master/ebook/zh/02.01.md
 
-def unwrap(x):
-    return -x
+  author: jiang
+  email:  [email protected]
 
+  created on 2014-5-25
+  """
+  return sorted(lst)[:k]
 
-def topk(lst, k):
-    h = []
-    for i in lst[0:k]:
-        heapq.heappush(h, wrap(i))
-    for i in lst[k:]:
-        heapq.heappushpop(h, wrap(i))
-    return map(unwrap, h)
+
+def ksmallest2(lst, k):
+  """
+  code example for section 2.1 solution 2
+  see https://github.com/julycoding/The-Art-Of-Programming-By-July/blob/master/ebook/zh/02.01.md
+
+  author: jiang
+  email:  [email protected]
+
+  created on 2014-5-25
+  """
+  result = lst[:k]
+  kmax = max(result)
+  for elem in lst[k+1:]:
+    if elem < kmax:
+      result.remove(kmax)
+      result.append(elem)
+      kmax = max(result)
+  return result
+
+
+def ksmallest3(lst, k):
+  """
+  code example for section 2.1 solution 3
+  see https://github.com/julycoding/The-Art-Of-Programming-By-July/blob/master/ebook/zh/02.01.md
+
+  modified by jiang
+  email   [email protected]
+
+  modified on 2014-5-24
+  """
+  heap = []
+  # 需要map是因为标准库中实现的是最小堆
+  tmp = map(lambda x: -x, lst)
+  for elem in tmp[:k]:
+    heappush(heap, elem)
+  for elem in tmp[k:]:
+    heappushpop(heap, elem)
+  return map(lambda x: -x, heap)
 
 
 if __name__ == "__main__":
-    l = [1, 9, 2, 4, 7, 6, 3]
+    lst = [1, 9, 2, 4, 7, 6, 3]
     k = 3
-    print topk(l, k)
+    print(ksmallest1(lst,k))
+    print(ksmallest2(lst,k))
+    print(ksmallest3(lst, k))
+