关于return语句的诡异事件

Question

def search(num,seq=[]):    if len(seq)==1:  #      print("3")#        assert seq[0]==num  #       num1=seq[0]         return seq[0]    elif num<seq[len(seq)//2]: #       print("2")         seq=seq[0:len(seq)//2]         search(num,seq=seq[:])        return 2    elif num > seq[len(seq)//2]:  #      print("1")         seq=seq[len(seq)//2+1:]         search(num,seq=seq[:])    else:   #     print("5")         return seq[len(seq)//2]def test(seq):    return 2seq=[1,2,3,4,5,6,7,8] seq.sort()print(search(6,seq=seq[:]))print(seq)print(test(seq))None [1, 2, 3, 4, 5, 6, 7, 8]2上例为实现二分法的示例，输出search(6,seq)函数应该返回搜索值，但是总是返回None,请各位帮忙解决

Answer 1

至少也应该是这样的：

def search(num,seq=[]):    if len(seq)==1:        return seq[0]    elif num<seq[len(seq)//2]:
        seq=seq[0:len(seq)//2]        return search(num,seq=seq[:])    elif num > seq[len(seq)//2]:
        seq=seq[len(seq)//2+1:]        return search(num,seq=seq[:])    else:        return seq[len(seq)//2]def test(seq):    return 2seq=[1,1,2,3,4,5,6,7,8]
seq.sort()print(search(6,seq=seq[:]))print(seq)print(test(seq))

 

但你的代码是返回搜索值（而非index），在不破坏源程序结构的情况下可以改成这样来返回index:

 

def search(num,seq=[]):
    mid=len(seq)//2    if len(seq)==1:        return 0    elif num<seq[mid]:        return mid+search(num,seq=seq[0:mid])    elif num > seq[mid]:        return mid+search(num,seq=seq[mid+1:])    else:        return middef test(seq):    return 2seq=[1,1,2,3,4,5,6,7,8]
seq.sort()print(search(6,seq=seq[:]))print(seq)print(test(seq))