题目如下:
Given a binary string
s
and an integerk
.Return True if every binary code of length
k
is a substring ofs
. 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: trueExample 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: falseConstraints:
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