python-二分查找

1.while循环实现

li = [1,4,6,8,9,10,15,18]
left = 0    
right = len(li)-1
n = 99
while 1:
    if left<right:
        mid = (left+right)//2
        if n>li[mid]:
            left = mid + 1
        if n <li[mid]:
            right = mid-1
        if n==li[mid]:
            print("找到了")
            break
    else:
        print("找不到")
        break

2.递归实现

li = [1,4,6,8,9,10,15,18]
def func(n,left,right):
    if left<=right:
        mid = (left+right)//2
        if n<li[mid]:
            right = mid-1
            return func(n,left,right)  # 但你用了递归后要返回一个值，前面的所有都要返回，不然是返回不到的
        if n>li[mid]:
            left = mid + 1
            return func(n,left,right)
        if n==li[mid]:
            print("找到了")
            return mid   # 返回索引
    else:
        print("找不到")
        return -1
a =func(4,0,len(li)-1)
print(a)

3.另类方法实现，缺点（无法准确计算出第n个数原列表的索引）

lst = [1,2,5,7,9,25,35,43,47,49,55,58,61]
def func(lst,taget):
    left = 0
    right = len(lst)-1
    if left>right:
        print("找不到")
    else:
        mid = (left+right)//2
        if taget>lst[mid]:

            return func(lst[mid+1:],taget)
        elif taget<lst[mid]:
            return func(lst[:mid],taget)
        else:
            print("找到了")
            return

func(lst,56)

 

来源：https://www.cnblogs.com/whileke/p/11318023.html