Python二分查找详解_Python

先来看个实例

									#!/usr/bin/env python 

									import sys  

									def search2(a,m): 

									  low = 0

									  high = len(a) - 1

									  while(low <= high): 

									    mid = (low + high)/2

									    midval = a[mid] 

									    if midval < m: 

									      low = mid + 1

									    elif midval > m: 

									      high = mid - 1

									    else: 

									      print mid  

									      return mid  

									  print -1

									  return -1

									if __name__ == "__main__": 

									  a = [int(i) for i in list(sys.argv[1])] 

									  m = int(sys.argv[2]) 

									  search2(a,m)

运行:

1	`administrator@ubuntu:~/Python$ python test_search2.py` `123456789` `4`

注:

1.'__'：由于python的类成员都是公有、公开的被存取public，缺少像正统面向对象语言的私有private属性。
于是就用__来将就一下，模拟私有属性。这些__属性往往是内部使用，通常情况下不用改写。也不用读取。
加上2个下划线的目的，一是不和普通公有属性重名冲突，二是不让对象的使用者(非开发者)随意使用。
2.__name__ == "__main__"表示程序脚本是直接被执行的.
如果不等于表示脚本是被其他程序用import引入的.则其__name__属性被设为模块名

Python采用二分查找找出数字的下标

要考虑有重复数字的情况

									class Solution(object):

									  def searchRange(self, nums, target):

									    """

									    :type nums: List[int]

									    :type target: int

									    :rtype: List[int]

									    """

									    def binary_search(start,end,value):

									      while end>=start:

									        mid = (start+end)//2

									        print(mid)

									        if nums[mid]>target:

									          end = mid-1

									        elif nums[mid]<target:

									          start = mid+1

									        else: 

									          if value==-1:

									            if mid-1>=start and nums[mid+value] == target:

									              end = mid+value

									            else:

									              return mid

									          else:

									            if mid+1<=end and nums[mid+value] == target:

									              start = mid+value

									            else:

									              return mid

									      return -1

									    a=binary_search(0,len(nums)-1,-1)

									    b=binary_search(0,len(nums)-1,1)

									    return [a,b]

									a = Solution()

									l = [2,2]

									print(a.searchRange(l,2))

二分算法的定义不在多说了，百度一下就知道（支持国产大笑）

									import sys 

									source = [1,2,3,4,5,6,7,8,9,10] #must be in order 

									des = int(sys.argv[1]) 

									low = 0

									high = len(source) - 1

									targetIndex = -1

									print "des=",des 

									while low <= high: 

									  middle = (low + high)/2

									  if des == source[middle]: 

									    targetIndex = middle 

									    break

									  elif des < source[middle]: 

									    high = middle -1

									    print "middle element[index=",middle,",value=",source[middle],"] is bigger than des, continue search from[",low,"to",high,"]"

									  else: 

									    low = middle + 1

									    print "middle element[index=",middle,",value=",source[middle],"] is smaller than des, continue search from[",low,"to",high,"]"

									print "search complete, target element's index in source list is ",targetIndex

最后在分享一个

									'fileName--BinarySearch.py'

									src = [] 

									def BinarySearch(low, high, target, *src): 

									  '二分查找'

									  while low <= high: 

									    mid = (low + high) // 2

									    midVal = src[mid] 

									    if target < midVal: 

									      high = mid - 1

									    elif target > midVal: 

									      low = mid + 1

									    else: 

									      return mid 

									    BinarySearch(low, high, target, *src) 

									print('Please input 10 number:') 

									for number in range(10): 

									  src.append(int(input('Num %d:' % number))) 

									sortList = tuple(src) 

									key = int(input('Please input key:')) 

									location = BinarySearch(0, len(src) - 1, key, *sortList) 

									if location != None: 

									  print('Find target at %d' % (location + 1)) 

									else: 

									  print('No target!')

Python二分查找详解

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