python一个关于二分法查找元素的实现

Posted 2020-12-30 sssantiago

# coding=utf-8
import time


def find_ele(alist, ele):
    if ele < alist[0] or ele > alist[len(alist) - 1]:
        print("%d not in alist" % ele)
        return
    last_index = len(alist) - 1
    center_index = last_index // 2
    loop_flag = True
    loop_cout = 0
    while loop_flag:
        # 使用二分法查找元素是否在有序列表内
        loop_cout += 1
        while alist[center_index] < ele:
            loop_cout += 1
            #最后一个元素获取
            center_index = (center_index + last_index + 1) // 2
            if center_index == last_index:
                # 最后一个元素
                if alist[center_index] == ele:
                    print("%d in alist" % ele)
                else:
                    print("%d not in alist" % ele)
                loop_flag = False
                break
        while alist[center_index] > ele:
            loop_cout += 1
            last_index = center_index
            center_index = last_index // 2
            if center_index == last_index:
                # 最后一个元素
                if alist[center_index] == ele:
                    print("%d in alist" % ele)
                else:
                    print("%d not in alist" % ele)
                loop_flag = False
                break
        if alist[center_index] == ele:
            print("%d in alist" % ele)
            loop_flag=False


def main():
    print(time.time())
    alist = list(range(100000000))
    # alist.pop(98)
    find_ele(alist, 100)
    print (time.time())

if __name__ == "__main__":
    main()