problem:https://leetcode.com/problems/binary-string-with-substrings-representing-1-to-n/
暴力枚举了一下所有子字符串可能对应的数字,统计是否达到N个,就过了。不知道有没有更快的方法。
class Solution { public:bool queryString(string S, int N) {unordered_set<int> count;for(int i = 0;i < S.size();i++){if(S[i] == '0') continue;int num = 0;for(int j = i;j < S.size();j++){num = 2 * num + S[j] - '0';if(num <= N){count.insert(num);}else{break;}}}return count.size() == N;} };
转载于:https://www.cnblogs.com/fish1996/p/11295487.html