Python编程中归并排序算法的实现步骤详解
基本思想：归并排序是一种典型的分治思想，把一个无序列表一分为二，对每个子序列再一分为二，继续下去，直到无法再进行划分为止。然后，就开始合并的过程，对每个子序列和另外一个子序列的元素进行比较，依次把小元素放入结果序列中进行合并，最终完成归并排序。
归并操作过程：
申请空间，使其大小为两个已经排序序列之和，该空间用来存放合并后的序列
设定两个指针，最初位置分别为两个已经排序序列的起始位置
比较两个指针所指向的元素，选择相对小的元素放入到合并空间，并移动指针到下一位置
重复步骤3直到某一指针达到序列尾
将另一序列剩下的所有元素直接复制到合并序列尾
上述说法是理论表述，下面用一个实际例子说明：
例如一个无序数组
[6,2,3,1,7]
首先将这个数组通过递归方式进行分解，直到：
[6],[2],[3],[1],[7]
然后开始合并排序，也是用递归的方式进行：
两个两个合并排序，得到：
[2,6],[1,3],[7]
上一步中，其实也是按照本步骤的方式合并的，只不过由于每个list中一个数，不能完全显示过程。下面则可以完全显示过程。
初始：
a = [2,6] b = [1,3] c = []
第1步，顺序从a,b中取出一个数字：2,1 比较大小后放入c中，并将该数字从原list中删除，结果是：
a = [2,6] b = [3] c = [1]
第2步，继续从a,b中按照顺序取出数字，也就是重复上面步骤，这次是：2,3 比较大小后放入c中，并将该数字从原list中删除，结果是：
a = [6] b = [3] c = [1,2]
第3步,再重复前边的步骤,结果是：
a = [6] b = [] c = [1,2,3]
最后一步，将6追加到c中，结果形成了：
a = [] b = [] c = [1,2,3,6]
通过反复应用上面的流程，实现[1,2,3,6]与[7]的合并
最终得到排序结果
[1,2,3,6,7]
本文列举了三种python的实现方法：
方法1：将前面讲述的过程翻译过来了，略先拙笨
#! /usr/bin/env python #coding:utf-8 def merge_sort(seq): if len(seq) ==1: return seq else: middle = len(seq)/2 left = merge_sort(seq[:middle]) right = merge_sort(seq[middle:]) i = 0 #left 计数 j = 0 #right 计数 k = 0 #总计数 while i < len(left) and j < len(right): if left[i] < right [j]: seq[k] = left[i] i +=1 k +=1 else: seq[k] = right[j] j +=1 k +=1 remain = left if i<j else right r = i if remain ==left else j while r<len(remain): seq[k] = remain[r] r +=1 k +=1 return seq
方法2：在按照顺序取数值方面，应用了list.pop()方法，代码更紧凑简洁
#! /usr/bin/env python #coding:utf-8 def merge_sort(lst): #此方法来自维基百科 if len(lst) <= 1: return lst def merge(left, right): merged = [] while left and right: merged.append(left.pop(0) if left[0] <= right[0] else right.pop(0)) while left: merged.append(left.pop(0)) while right: merged.append(right.pop(0)) return merged middle = int(len(lst) / 2) left = merge_sort(lst[:middle]) right = merge_sort(lst[middle:]) return merge(left, right)
方法3：原来在python的模块heapq中就提供了归并排序的方法，只要将分解后的结果导入该方法即可。
#! /usr/bin/env python #coding:utf-8 from heapq import merge def merge_sort(seq): if len(seq) <= 1: return m else: middle = len(seq)/2 left = merge_sort(seq[:middle]) right = merge_sort(seq[middle:]) return list(merge(left, right)) #heapq.merge() if __name__=="__main__": seq = [1,3,6,2,4] print merge_sort(seq)
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