[Leetcode 1018] Binary Prefix Divisible By 5

原题说明

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

解题思路

题目要求检查从A[0] 到 A[i] (0 <= i < A.size())的字符串，对应的二进制数是否能被5整除。

我们按顺序依次检查二进制对应的数是否能被5整除。

这里有一个小技巧：两个数的和对5的余数，等于它们各自对5的余数的和对5的余数。运用这个事实，我们可以防止整数溢出。

示例代码 (cpp)

class Solution {
public:
    vector<bool> prefixesDivBy5(vector<int>& A) {
        vector<bool> ans;
        int tmp = 0;
        for (auto a : A) {
            tmp = (tmp * 2 + a) % 5;
            if (tmp == 0) {
                ans.push_back(true);
            }
            else {
                ans.push_back(false);
            }
        }
        return ans;
    }
};

复杂度分析

时间复杂度: O(N)
空间复杂度: O(N)

归纳总结

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