1018. Binary Prefix Divisible By

题目描述

Given an array A of 0s and 1s, consider N_i: the i-th subarray from A[0] to A[i] interpreted as a binary number (from most-significant-bit to least-significant-bit.)

Return a list of booleans answer, where answer[i] is true if and only if N_i is divisible by 5.

Example 1:

Input: [0,1,1]
Output: [true,false,false]
Explanation: 
The input numbers in binary are 0, 01, 011; which are 0, 1, and 3 in base-10.  Only the first number is divisible by 5, so answer[0] is true.

Example 2:

Input: [1,1,1]
Output: [false,false,false]

Example 3:

Input: [0,1,1,1,1,1]
Output: [true,false,false,false,true,false]

Example 4:

Input: [1,1,1,0,1]
Output: [false,false,false,false,false]

Note:

1 <= A.length <= 30000
A[i] is 0 or 1

解题思路

最开始的时候的思路是：

• 每一次都划分成一个切片
• 然后将切片，也就是二进制数组转换为相应的十进制数字
• 然后对转换后的十进制数字进行判断
  但是在实际的写的过程中，发现切片的动态调整要求很多（可能是个人原因，一会是数组越界，一会是切片无法定义）之后突然发现，其实并不需要这么麻烦，我们只需要将num乘以2之后再加上当前对应的数组的内容A[i]就可以了~~

具体实现

func prefixesDivBy5(A []int) []bool {
    num, rlt := 0, make([]bool, len(A))
    for i := 0; i < len(A); i++ {
        num = (num * 2 + A[i])  
        if num % 5 == 0 {
            rlt[i] = true
        }
    }
    
    return rlt
}