Python Binary Search details, python binary explanation

Python Binary Search details, python binary explanation

Let's first look at an instance.

#!/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) 

Run:

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

3

Note:

1. '_': Because python class members are both public and publicly accessible and public, there is a lack of private attributes like the Orthodox object-oriented language.
So we will use _ to simulate private attributes. These _ attributes are often used internally and generally do not need to be rewritten. No need to read.
The purpose of adding two underscores (_) is not to conflict with the names of common public properties, or to prevent users (non-developers) of objects from using them at will.
2. _ name _ = "_ main _" indicates that the script is directly executed.
If the value is not equal to, the script is imported by other programs using import. The _ name _ attribute is set as the module name.

Python uses binary search to find the subscript of a number

Consider repeated numbers.

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))

The definition of the binary algorithm is not too much. Baidu will know it at a Glance (supporting domestic laugh)

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 

Finally, share

'Filename -- BinarySearch. py 'src = [] def BinarySearch (low, high, target, * src): 'binary look' 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! ')