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
}
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