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


    题目如下:

    Given a binary string s and an integer k.

    Return True if every binary code 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

    解题思路:本题采用逆向思维,先计算出s中长度为k的子串,去重后判断总数是否为2^k即可。

    代码如下:

    class Solution(object):
        def hasAllCodes(self, s, k):
            """
            :type s: str
            :type k: int
            :rtype: bool
            """
            dic = {}
            subs = ''
            for i in range(len(s)):
                if len(subs) < k:
                    subs += s[i]
                else:
                    subs = subs[1:] + s[i]
                if len(subs) == k:
                    dic[subs] = 1
            return len(dic) == 2**k
    
            
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  • 原文地址:https://www.cnblogs.com/seyjs/p/13111728.html
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