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 <= S.length <= 1000
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; } };