标签:pre NPU bin The turn res public exist lang
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
判断长度为k的二进制0/1的所有组合是否都是s的子串。
将s中每一个长度为k的子串都加入到set中,最后判断set的容量是否是2^k。
class Solution {
public boolean hasAllCodes(String s, int k) {
Set<String> set = new HashSet<>();
for (int i = 0; i + k <= s.length(); i++) {
set.add(s.substring(i, i + k));
if (set.size() == (1 << k)) return true;
}
return false;
}
}
1461. Check If a String Contains All Binary Codes of Size K (M)
标签:pre NPU bin The turn res public exist lang
原文地址:https://www.cnblogs.com/mapoos/p/14525910.html