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  • 1016. Binary String With Substrings Representing 1 To N

    Given a binary string S (a string consisting only of '0' and '1's) and a positive integer N, return true if and only if for every integer X from 1 to N, the binary representation of X is a substring of S.

    Example 1:

    Input: S = "0110", N = 3
    Output: true
    

    Example 2:

    Input: S = "0110", N = 4
    Output: false
    

    Note:

    1. 1 <= S.length <= 1000
    2. 1 <= N <= 10^9

    这个应该算暴力解法???

    优化点: 由于是判断二进制是否存在,我们知道二进制左移一位等于除以2,所以如果N存在,那么N/2肯定也存在, 同理N/4,N/8,N/16都是的.

    所以判断1到N是否都是子串,只要判断N到(N/2-1) 即可. N/2本身不需要判断 

    class Solution {
    public:
        bool queryString(string S, int N) {
            for(int i=N;i>N/2;--i)
            {
                string bin=bitset<32>(i).to_string();
                if(string::npos==S.find(bin.substr(bin.find("1"))))
                    return false;
            }
            return true;
        }
    };
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  • 原文地址:https://www.cnblogs.com/lychnis/p/11973922.html
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