怎么从Ruby数组中删除元素

Python015

怎么从Ruby数组中删除元素,第1张

irb(main):001:0>a=[1,2,5,3,9,0,9]

=>[1, 2, 5, 3, 9, 0, 9]

irb(main):002:0>a-[2]

=>[1, 5, 3, 9, 0, 9]

irb(main):004:0>a

=>[1, 2, 5, 3, 9, 0, 9]

irb(main):005:0>a-[4]

=>[1, 2, 5, 3, 9, 0, 9]

public static void main(String[] args)

{

int [] num =new int[]{1,2,3,4,5,6,7,8,9,10}

Scanner input = new Scanner(System.in)

int temp=0

System.out.println("请输入你要删除的元素:")

int de=input.nextInt()

for(int i =0i<num.lengthi++)

{

if(num[i]==de)

{

for(int j = ij<num.length-1j++)

{

num[j]=num[j+1]

}

temp=1

break

}

}

if(temp==1)

{

System.out.println("删除此对象后数组值为：")

for(int i = 0i<num.length-1i++)

{

System.out.print(num[i]+"\t")

}

}

else

{

System.out.println("未找到你要删除的对象")

}

}

先把字符串转成字符数组，然后写个函数扫一遍字符数组，遇到2的倍数步就加个空格，返回新字符串

字符串转成字符数组：

>>str = "ABC"

=>"ABC"

>>chars = str.scan(/./)

=>["A", "B", "C"]

函数给你个python的参考吧，话说我python居然还真没忘。。。

'''正向操作

def f(x):

s = ""

for i in range(len(x)):

if i % 2 == 0:

s = s + ' '

s = s + x[i]

return s[1:]

'''逆向操作

def f_rev(x):

s = ""

'''凡是偶数位的空格删掉

'''

for i in range(len(x)):

if (i-2) % 3 == 0 and x[i] == ' ':

continue

s += x[i]

return s

'''测试

s = "abcdedg"

print (f_rev(f(s))