題目描述
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
解題思路
最開始的時候的思路是：
- 每一次都劃分成一個切片
- 然后將切片绸栅，也就是二進(jìn)制數(shù)組轉(zhuǎn)換為相應(yīng)的十進(jìn)制數(shù)字
- 然后對轉(zhuǎn)換后的十進(jìn)制數(shù)字進(jìn)行判斷
但是在實際的寫的過程中担神，發(fā)現(xiàn)切片的動態(tài)調(diào)整要求很多（可能是個人原因懦尝，一會是數(shù)組越界忆绰，一會是切片無法定義）之后突然發(fā)現(xiàn)，其實并不需要這么麻煩速侈，我們只需要將num乘以2之后再加上當(dāng)前對應(yīng)的數(shù)組的內(nèi)容A[i]就可以了~~
具體實現(xiàn)
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
}
