M1461.检查一个字符串是否包含所有长度为 K 的二进制子串

bit manipulation, https://leetcode.cn/problems/check-if-a-string-contains-all-binary-codes-of-size-k/

思路：滑动窗口, 学会了 stoi (string to int) 的用法, int stoi( const std::string& str, std::size_t* pos = nullptr, int base = 10 )

#include <bits/stdc++.h>
using namespace std;

class Solution {
public:
    bool hasAllCodes(string s, int k) {
        int num = stoi(s.substr(0, k), nullptr, 2);
        unordered_set<int> a;
        a.insert(num);
        for (int i = k; i < s.length(); i++) {
            num <<= 1;
            if (num >= (1 << k)) {
                num -= (1 << k);
            }
            num += (s[i] - '0');
            a.insert(num);
        }

        return a.size() == (1 << k);
    }
};

用时10min

#include <iostream>
#include <vector>
#include <unordered_set>
using namespace std;

class Solution
{
public:
    bool hasAllCodes(string s, int k)
    {
        int total_needed = 1 << k;
        if (s.size() < (size_t)total_needed + k - 1)
            return false;

        vector<bool> seen(total_needed, false);

        int x = 0;
        int cnt = 0;
        int MASK = total_needed - 1;

        for (int i = 0; i < k - 1; ++i)
        {
            x = (x << 1) | (s[i] & 1);
        }

        for (int i = k - 1; i < s.length(); ++i)
        {
            x = ((x << 1) & MASK) | (s[i] & 1);
            if (!seen[x])
            {
                seen[x] = true;
                if (++cnt == total_needed)
                    return true;
            }
        }

        return false;
    }
};

int main()
{
    cin.tie(nullptr)->sync_with_stdio(false);

    Solution sol;
    string s = "0110";
    int k = 2;
    cout << sol.hasAllCodes(s, k) << '\n';
    return 0;
}

共用时20min

思路：直接提取 $s$ 的所有长度为 $k$ 的子串．时间复杂度为 $O(|s|)$．

class Solution {
   public:
    bool hasAllCodes(string s, int k) {
        const int n = s.length();
        if(n - k + 1 < (1 << k)) return false;
        vector<bool> buc;
        buc.resize(1 << k);
        int x = 0;
        for(int i = 0; i < n; i++) {
            x <<= 1;
            x |= s[i] - '0';
            x &= (1 << k) - 1;
            if(i >= k - 1) buc[x] = true;
        }
        return ranges::count(buc, false) == 0;
    }
};