1018. Binary Prefix Divisible by 5

1018. Binary Prefix Divisible By 5 #

题目 #

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

题目大意 #

给定由若干 0 和 1 组成的数组 A。我们定义 N_i:从 A[0] 到 A[i] 的第 i 个子数组被解释为一个二进制数(从最高有效位到最低有效位)。返回布尔值列表 answer,只有当 N_i 可以被 5 整除时,答案 answer[i] 为 true,否则为 false。

解题思路 #

  • 简单题。每扫描数组中的一个数字,累计转换成二进制数对 5 取余,如果余数为 0,则存入 true,否则存入 false。

代码 #

package leetcode

func prefixesDivBy5(a []int) []bool {
	res, num := make([]bool, len(a)), 0
	for i, v := range a {
		num = (num<<1 | v) % 5
		res[i] = num == 0
	}
	return res
}

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