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

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

题目 #

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

题目大意 #

给你一个二进制字符串 s 和一个整数 k 。如果所有长度为 k 的二进制字符串都是 s 的子串,请返回 True ,否则请返回 False

解题思路 #

  • 构造一个 mask 遮罩,依次划过整个二进制字符串,每次滑动即取出遮罩遮住的 k 位二进制字符。可以用 map 存储不同的二进制转换成的十进制数,最后判断 len(map) 是否等于 k 即可。但是用 map 存比较慢,此处换成 bool 数组。先构造一个长度为 k 的数组,然后每次通过 mask 更新这个 bool 数组对应十进制的 bool 值,并且记录剩余还缺几个二进制数。等剩余的等于 0 的时候,说明所有二进制字符串都出现了,直接输出 true,否则循环完以后输出 false

代码 #

package leetcode

import "math"

func hasAllCodes(s string, k int) bool {
	need := int(math.Pow(2.0, float64(k)))
	visited, mask, curr := make([]bool, need), (1<<k)-1, 0
	for i := 0; i < len(s); i++ {
		curr = ((curr << 1) | int(s[i]-'0')) & mask
		if i >= k-1 { // mask 有效位达到了 k 位
			if !visited[curr] {
				need--
				visited[curr] = true
				if need == 0 {
					return true
				}
			}
		}
	}
	return false
}

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