【LeetCode】1018. Binary Prefix Divisible By 5 可被 5 整除的二进制前缀(Easy)(JAVA)
题目地址: https://leetcode.com/problems/binary-prefix-divisible-by-5/
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
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。
解题方法
- 就是计算前 n 位构成的二进制数是否可以被 5 整除
- 已经知道前 n 位的结果 pre, 现在多了 n + 1 位,怎么求出结果呢? 就是前 n 位统一往左移一位即可 pre << 1, 再加上 A[n + 1] 的结果
- note: 为了防止 int 超限需要对求出的结果 pre 进行取余数: pre % 5
class Solution {
public List<Boolean> prefixesDivBy5(int[] A) {
List<Boolean> res = new ArrayList<>();
int pre = 0;
for (int i = 0; i < A.length; i++) {
pre <<= 1;
pre += A[i];
pre = pre % 5;
res.add(pre == 0);
}
return res;
}
}
执行耗时:4 ms,击败了92.76% 的Java用户
内存消耗:39.2 MB,击败了41.06% 的Java用户