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

    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);
        }
    }

    看了hint我尼玛狂躁,不过想想也是,只要长度是k的substring恰好有2^k个就行了,差一点啊差一点

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  • 原文地址:https://www.cnblogs.com/wentiliangkaihua/p/13029413.html
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