Leetcode 1029. Binary Prefix Divisible By 5

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

方法一：利用大数整数

public List<Boolean> prefixesDivBy5(int[] A) {
        ArrayList<Boolean> res = new ArrayList<>(A.length);
        BigInteger bInt = BigInteger.ZERO;
        for (int i = 0; i < A.length; i++) {
            bInt = bInt.shiftLeft(1).add(A[i] == 1 ? BigInteger.ONE : BigInteger.ZERO);
            res.add(bInt.mod(BigInteger.valueOf(5l)) == BigInteger.ZERO);
        }
        return res;
    }

方法二：利用余数规律

public static List<Boolean> prefixesDivBy5(int[] A) {
        ArrayList<Boolean> res = new ArrayList<>(A.length);
        // pair's key is the reminder when meet '0';
        // pair's val is the reminder when meet '1'.
        // stateDict's index(0,1,2,3,4) is the possible reminder
        Pair<Integer, Integer>[] stateDic = new Pair[]{
                new Pair<>(0,1),
                new Pair<>(2,3),
                new Pair<>(4,0),
                new Pair<>(1,2),
                new Pair<>(3,4),
        };
        int state = 0;
        for (int i = 0; i < A.length; i++) {
            state = A[i] == 0 ? stateDic[state].getKey() : stateDic[state].getValue();
            res.add(state == 0);
        }
        return res;
    }