二分查找法

# -*- coding:utf-8 -*-

# 写一下二分查找，时间复杂度为O(logn), n为查询元素的个数


l1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]


# 二分查找
def bisearch(l, find_obj):
    length = len(l)
    find_offect = length // 2
    print(find_offect)
    if find_obj > l[find_offect]:
        ret = bisearch(l[find_offect:], find_obj)
        return ret
    elif find_obj < l[find_offect]:
        ret = bisearch(l[:find_offect-1], find_obj)
        return ret
    else:
        return l[find_offect]


print(bisearch(l1, 10))

# 原始方法
for i in l1:
    print(i)
    if i == 10:
        print('%s' % i)
        exit(0)


5
3
1
10
1
2
3
4
5
6
7
8
9
10
10