package com.example.lettcode.dailyexercises;
import java.util.ArrayList;
import java.util.List;
/**
* @Class PrefixesDivBy5
* @Description 1018 可被5整除的二进制前缀
* 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。
* <p>
* 示例 1:
* 输入:[0,1,1]
* 输出:[true,false,false]
* 解释:
* 输入数字为 0, 01, 011;也就是十进制中的 0, 1, 3 。只有第一个数可以被 5 整除,因此 answer[0] 为真。
* <p>
* 示例 2:
* 输入:[1,1,1]
* 输出:[false,false,false]
* <p>
* 示例 3:
* 输入:[0,1,1,1,1,1]
* 输出:[true,false,false,false,true,false]
* <p>
* 示例 4:
* 输入:[1,1,1,0,1]
* 输出:[false,false,false,false,false]
* <p>
* 提示:
* 1 <= A.length <= 30000
* A[i] 为 0 或 1
* @Author
* @Date 2021/1/14
**/
public class PrefixesDivBy5 {
public static List<Boolean> prefixesDivBy5(int[] A) {
if (A == null || A.length == 0) return new ArrayList<>();
List<Boolean> booleanList = new ArrayList<>();
int count = A[0];
for (int i = 1; i < A.length; i++) {
if (count % 5 == 0) booleanList.add(true);
else booleanList.add(false);
count = (count * 2 + A[i]) % 10;
}
if (count % 5 == 0) booleanList.add(true);
else booleanList.add(false);
return booleanList;
}
}
// 测试用例
public static void main(String[] args) {
int[] A = new int[]{0, 1, 1};
List<Boolean> booleanList = prefixesDivBy5(A);
System.out.print("PrefixesDivBy5 demo01 result:[");
for (Boolean ble : booleanList) {
System.out.print(" " + ble);
}
System.out.println("]");
A = new int[]{1, 1, 1};
booleanList = prefixesDivBy5(A);
System.out.print("PrefixesDivBy5 demo02 result:[");
for (Boolean ble : booleanList) {
System.out.print(" " + ble);
}
System.out.println("]");
A = new int[]{0, 1, 1, 1, 1, 1};
booleanList = prefixesDivBy5(A);
System.out.print("PrefixesDivBy5 demo03 result:[");
for (Boolean ble : booleanList) {
System.out.print(" " + ble);
}
System.out.println("]");
A = new int[]{1, 1, 1, 0, 1};
booleanList = prefixesDivBy5(A);
System.out.print("PrefixesDivBy5 demo04 result:[");
for (Boolean ble : booleanList) {
System.out.print(" " + ble);
}
System.out.println("]");
}