binary_search 二分查找

1.在列表中获取中间位置的值

2.将中间值和所需要查找的值做对比, 如果相等则返回中间值的位置

3.如果中间值小于所需要查找的值, 则将查找范围缩小到 left 至 mid-1 (即right 修改为 mid-1)

4.如果中间值大于所需要查找的值, 则将查找范围修改为 mid+1 至 right (即left 修改为 mid+1)

注: 该算法 永远只查找 中间值是否等于查找值, 无法找到第一个出现 或者最后一个出现的位置

 

import numpy as np
a = list(np.random.randint(low=0,high=10,size=10))
# sort
a.sort()
print(a)

# a = [0, 2, 2, 3, 4, 7, 7, 7, 8, 9]
a = [0, 3, 3, 4, 4, 5, 6, 8, 9, 9]
def search_b(num):
    # set left and right
    left = 0
    right = len(a)-1
    
    while left < right:
        # print("left:%d, right:%d"%(left, right))
        # set mid
        mid = (right +left)//2
        # print("mid:%d"%(mid))
        # print(a[mid])
        if a[mid] == num:
            print("The Number:%d is located %d th in list"%(num, mid))
            break
        elif a[mid] < num:
            left = mid+1
            # print("mid < num left:%d;right%d"%(left, right))
        else:
            right =mid-1
            # print("mid > num left:%d;right%d"%(left, right))
        
    return -1

print(search_b(3))