二分查找---二分法

Posted 2020-10-06

定义：二分查找又称折半查找，优点是比较次数少，查找速度快，平均性能好;其缺点是要求待查表为有序表，且插入删除困难。因此，折半查找方法适用于不经常变动而查找频繁的有序列表。首先，假设表中元素是按升序排列，将表中间位置记录的关键字与查找关键字比较，如果两者相等，则查找成功;否则利用中间位置记录将表分成前、后两个子表，如果中间位置记录的关键字大于查找关键字，则进一步查找前一子表，否则进一步查找后一子表。重复以上过程，直到找到满足条件的记录，使查找成功，或直到子表不存在为止，此时查找不成功

#Author wangmengzhu
def binary_search(lst,item):
    low = 0
    high = len(lst) -1
    while low <= high:
        mid = int((low + high)/2)
        guess = lst[mid]
        if guess == item:
            return mid
        elif guess > item:
            high = mid - 1
        else:
            low = mid +1

print(binary_search(lst = [1,2,3,5,34,5,4,6,67],item = 5))

 

l = [1,2,5,7,10,31,44,47,56,99,102,130,240]
def search_binary(l,item):
    mid_index = len(l) // 2
    if len(l) == 1:
        if l[0] == item:
            print(‘find it‘)
        else:
            print(‘not exists‘)
    if len(l) > 1:
        if l[mid_index] > item:
            l = l[:mid_index]
            search_binary(l,item)
        elif l[mid_index] < item:
            l = l[mid_index:]
            search_binary(l,item)
        else:
            print(‘find it‘)
search_binary(l,32)

l = [1,2,5,7,10,31,44,47,56,99,102,130,240]
def search_binary(l,item):
    mid_index = len(l) // 2
if len(l) == 1:
        if l[0] == item:
            print(‘find it‘)
        else:
            print(‘not exists‘)
    if len(l) > 1:
        if l[mid_index] > item:
            l = l[:mid_index]
            search_binary(l,item)
        elif l[mid_index] < item:
            l = l[mid_index:]
            search_binary(l,item)
        else:
            print(‘find it‘)
search_binary(l,32)