1461. Check If a String Contains All Binary Codes of Size K

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mb5fd340813ba80 2020-06-02 09:45:00

文章标签 i++ it 文章分类 代码人生

Given a binary string s and an integer k.

Return True if all binary codes of length k is a substring of s. Otherwise, return False.

Example 1:

Input: s = "00110110", k = 2
Output: true
Explanation: The binary codes of length 2 are "00", "01", "10" and "11". They can be all found as substrings at indicies 0, 1, 3 and 2 respectively.

Example 2:

Input: s = "00110", k = 2
Output: true

Example 3:

Input: s = "0110", k = 1
Output: true
Explanation: The binary codes of length 1 are "0" and "1", it is clear that both exist as a substring.

Example 4:

Input: s = "0110", k = 2
Output: false
Explanation: The binary code "00" is of length 2 and doesn't exist in the array.

Example 5:

Input: s = "0000000001011100", k = 4
Output: false

Constraints:

1 <= s.length <= 5 * 10^5
s consists of 0's and 1's only.
1 <= k <= 20

class Solution {
    public boolean hasAllCodes(String s, int k) {
        List<String> bits = new ArrayList();
        for(int i = 0; i < s.length() - k + 1; i++){
            bits.add(s.substring(i, i + k));
        }
        List<String> kbits = new ArrayList();
        for(int i = 0; i < Math.pow(2, k); i++){
            String ss = Integer.toBinaryString(i);
            while(ss.length() < k){
                ss = "0"+ss;
            }
            if(bits.indexOf(ss) < 0){
                return false;
            }
            // StringBuilder sb = new StringBuilder(Integer.toBinaryString(i));
            // sb = sb.reverse();
            // while(sb.length() < k){
            //     sb.append("0");
            // }
            // kbits.add(sb.reverse().toString());
        }
        return true;
    }
}

brute force死活过不了最后一个test case，铁傻逼嗷

class Solution {
    public boolean hasAllCodes(String s, int k) {
        Set<String> bits = new HashSet();
        for(int i = 0; i < s.length() - k + 1; i++){
            bits.add(s.substring(i, i + k));
        }
        return bits.size() == Math.pow(2, k);
    }
}