[LeetCode] 1018. Binary Prefix Divisible By 5 可被5整除的二进制前缀

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

这道题给了一个只由0和1组成的数组，问从0开始每个子数组表示的二进制数是否可以整除5，二进制数是从高位到低位的。既然是一道 Easy 的题目，也就不用太多的技巧，直接按顺序遍历即可。首先对于第一个数字，可以快速知道其是否可以整除5，当子数组新加一位，实际上相当于之前的数字左移了一位，也就相当于乘以了2，所以新的子数组表示的数字就是之前的数字乘以2再加上新加进来的数字，然后就可以判断是否可以整除5了。但是需要注意的一点是，由于A数组可能会很长，所以最终累加出来的数字可能会很大，超过整型最大值，甚至也超过长整型的最大值，为了避免这种情况，对每次累加出来的新数字都对5取余，这样就不会溢出了，参见代码如下：

class Solution {
public:
    vector<bool> prefixesDivBy5(vector<int>& A) {
        vector<bool> res;
        int cur = 0, n = A.size();
        for (int i = 0; i < n; ++i) {
            cur = (cur * 2 + A[i]) % 5;
            res.push_back(cur % 5 == 0);
        }
        return res;
    }
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1018

参考资料：

https://leetcode.com/problems/binary-prefix-divisible-by-5/

https://leetcode.com/problems/binary-prefix-divisible-by-5/discuss/265601/Detailed-Explanation-using-Modular-Arithmetic-O(n)

https://leetcode.com/problems/binary-prefix-divisible-by-5/discuss/265554/JavaPython-3-71-liners-left-shift-bitwise-or-and-mod.

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