检查给定的字符串是否是 K 周期的
原文:https://www . geeksforgeeks . org/check-if-给定字符串-k-periodic/
给定一个字符串 str 和一个整数 K ，任务是检查给定的字符串是否为 K 周期。如果字符串是子字符串字符串【0…k-1】的重复，即字符串“ababab”是 2 周期，则字符串是 k 周期的。打印是如果给定字符串是 k 周期的，则打印否。
示例:
输入: str = "geeksgeeks "，k = 5 输出:是 重复长度为 k 的前缀即“geeks”即可生成给定字符串
输入: str = "geeksforgeeks "，k = 3 输出:否
方式:从子串str【k，2k-1】、str【2k，3k-1】等开始，检查这些子串是否都等于长度为 k 的串的前缀，即str【0，k-1】。如果所有这些子字符串的条件都为真，则打印是否则打印否。 以下是上述方法的实施:
C++
// CPP implementation of the approach
#include<bits/stdc++.h>
using namespace std;
// Function that return true if sub-string
// of length k starting at index i is also
// a prefix of the string
bool isPrefix(string str, int len, int i, int k)
{
// k length sub-string cannot start at index i
if (i + k > len)
return false;
for (int j = 0; j < k; j++)
{
// Character mismatch between the prefix
// and the sub-string starting at index i
if (str[i] != str[j])
return false;
i++;
}
return true;
}
// Function that returns true if str is K-periodic
bool isKPeriodic(string str, int len, int k)
{
// Check whether all the sub-strings
// str[0, k-1], str[k, 2k-1] ... are equal
// to the k length prefix of the string
for (int i = k; i < len; i += k)
if (!isPrefix(str, len, i, k))
return false;
return true;
}
// Driver code
int main()
{
string str = "geeksgeeks";
int len = str.length();
int k = 5;
if (isKPeriodic(str, len, k))
cout << ("Yes");
else
cout << ("No");
}
// This code is contributed by
// Surendra_Gangwar
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG {
// Function that return true if sub-string
// of length k starting at index i is also
// a prefix of the string
static boolean isPrefix(String str, int len, int i, int k)
{
// k length sub-string cannot start at index i
if (i + k > len)
return false;
for (int j = 0; j < k; j++) {
// Character mismatch between the prefix
// and the sub-string starting at index i
if (str.charAt(i) != str.charAt(j))
return false;
i++;
}
return true;
}
// Function that returns true if str is K-periodic
static boolean isKPeriodic(String str, int len, int k)
{
// Check whether all the sub-strings
// str[0, k-1], str[k, 2k-1] ... are equal
// to the k length prefix of the string
for (int i = k; i < len; i += k)
if (!isPrefix(str, len, i, k))
return false;
return true;
}
// Driver code
public static void main(String[] args)
{
String str = "geeksgeeks";
int len = str.length();
int k = 5;
if (isKPeriodic(str, len, k))
System.out.print("Yes");
else
System.out.print("No");
}
}
Python 3
# Python3 implementation of the approach
# Function that returns true if sub-string
# of length k starting at index i
# is also a prefix of the string
def isPrefix(string, length, i, k):
# k length sub-string cannot
# start at index i
if i + k > length:
return False
for j in range(0, k):
# Character mismatch between the prefix
# and the sub-string starting at index i
if string[i] != string[j]:
return False
i += 1
return True
# Function that returns true if
# str is K-periodic
def isKPeriodic(string, length, k):
# Check whether all the sub-strings
# str[0, k-1], str[k, 2k-1] ... are equal
# to the k length prefix of the string
for i in range(k, length, k):
if isPrefix(string, length, i, k) == False:
return False
return True
# Driver code
if __name__ == "__main__":
string = "geeksgeeks"
length = len(string)
k = 5
if isKPeriodic(string, length, k) == True:
print("Yes")
else:
print("No")
# This code is contributed
# by Rituraj Jain
C
// C# implementation of the approach
using System;
class GFG
{
// Function that return true if sub-string
// of length k starting at index i is also
// a prefix of the string
static bool isPrefix(String str, int len, int i, int k)
{
// k length sub-string cannot start at index i
if (i + k > len)
return false;
for (int j = 0; j < k; j++)
{
// Character mismatch between the prefix
// and the sub-string starting at index i
if (str[i] != str[j])
return false;
i++;
}
return true;
}
// Function that returns true if str is K-periodic
static bool isKPeriodic(String str, int len, int k)
{
// Check whether all the sub-strings
// str[0, k-1], str[k, 2k-1] ... are equal
// to the k length prefix of the string
for (int i = k; i < len; i += k)
if (!isPrefix(str, len, i, k))
return false;
return true;
}
// Driver code
public static void Main()
{
String str = "geeksgeeks";
int len = str.Length;
int k = 5;
if (isKPeriodic(str, len, k))
Console.Write("Yes");
else
Console.Write("No");
}
}
/* This code contributed by PrinciRaj1992 */
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function that return true if sub-
// of length $k starting at index $i
// is also a prefix of the string
function isPrefix($str, $len, $i, $k)
{
// $k length sub- cannot start at index $i
if ($i + $k > $len)
return false;
for ( $j = 0; $j < $k; $j++)
{
// Character mismatch between the prefix
// and the sub- starting at index $i
if ($str[$i] != $str[$j])
return false;
$i++;
}
return true;
}
// Function that returns true if $str is K-periodic
function isKPeriodic($str, $len, $k)
{
// Check whether all the sub-strings
// $str[0, $k-1], $str[$k, 2k-1] ... are equal
// to the $k length prefix of the
for ($i = $k; $i < $len; $i += $k)
if (!isPrefix($str, $len, $i, $k))
return false;
return true;
}
// Driver code
$str = "geeksgeeks";
$len = strlen($str);
$k = 5;
if (isKPeriodic($str, $len, $k))
echo ("Yes");
else
echo ("No");
// This code is contributed by ihritik
?>
java 描述语言
<script>
// js implementation of the approach
// Function that return true if sub-string
// of length k starting at index i is also
// a prefix of the string
function isPrefix(str, len, i, k){
// k length sub-string cannot start at index i
if (i + k > len)
return false;
for (let j = 0; j < k; j++)
{
// Character mismatch between the prefix
// and the sub-string starting at index i
if (str[i] != str[j])
return false;
i++;
}
return true;
}
// Function that returns true if str is K-periodic
function isKPeriodic(str, len, k)
{
// Check whether all the sub-strings
// str[0, k-1], str[k, 2k-1] ... are equal
// to the k length prefix of the string
for (let i = k; i < len; i += k)
if (!isPrefix(str, len, i, k))
return false;
return true;
}
// Driver code
let str = "geeksgeeks";
let len = str.length;
let k = 5;
if (isKPeriodic(str, len, k))
document.write("Yes");
else
document.write("No");
</script>
Output:
Yes
时间复杂度: O(K * log(len)) 辅助空间: O(1)
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