检查给定的字符串是否是线性的
给定字符串 str ，任务是检查给定字符串是否是线性的。如果是线性，则打印“是”否则打印“否”。
让字符串为“abcdefghij”。可以分解为: 【a】 【BC】 【def】 【ghij】 如果字符 a、b、c 和相等，那么给定的字符串是线性的，否则不是。 因此，如果字符串的形式为“xxabxabcxabcdxabdexab……”那么它被称为线性字符串。
示例:
输入:str = " aapaxayaziabcde " 输出:是 解释:T8】a AP axy ayzi abcde 所有断串的字符都与第一个字符相同。 输入: str = "bbobfycd" 输出:No T21【解释: b bo bfy CD 这里 b=b=b！=c
方法:要检查给定字符串的形式是否为“xxabxabcxabcdexaab……”那么我们必须检查索引 0、1、3、6、10、… 处的字符是否相等。 如果上述索引处的所有字符相等，则给定的字符串为线性字符串，否则不相等。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check if the given
// string is linear or not
int is_linear(string s)
{
int tmp = 0;
char first = s[0];
// Iterate over string
for (int pos = 0; pos < s.length();
pos += tmp) {
// If character is not same as
// the first character then
// return false
if (s[pos] != first) {
return false;
}
// Increment the tmp
tmp++;
}
return true;
}
// Driver Code
int main()
{
// Given String str
string str = "aapaxyayziabcde";
// Function Call
if (is_linear(str)) {
cout << "Yes" << endl;
}
else {
cout << "No" << endl;
}
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.util.*;
class GFG{
// Function to check if the given
// string is linear or not
static boolean is_linear(String s)
{
int tmp = 0;
char first = s.charAt(0);
// Iterate over string
for(int pos = 0; pos < s.length();
pos += tmp)
{
// If character is not same as
// the first character then
// return false
if (s.charAt(pos) != first)
{
return false;
}
// Increment the tmp
tmp++;
}
return true;
}
// Driver code
public static void main(String[] args)
{
// Given String str
String str = "aapaxyayziabcde";
// Function call
if (is_linear(str))
{
System.out.println("Yes");
}
else
{
System.out.println("No");
}
}
}
// This code is contributed by offbeat
Python 3
# Python3 program for the above approach
# Function to check if the given
# is linear or not
def is_linear(s):
tmp = 0
first = s[0]
pos = 0
# Iterate over string
while pos < len(s):
# If character is not same as
# the first character then
# return false
if (s[pos] != first):
return False
# Increment the tmp
tmp += 1
pos += tmp
return True
# Driver Code
if __name__ == '__main__':
# Given String str
str = "aapaxyayziabcde"
# Function call
if (is_linear(str)):
print("Yes")
else:
print("No")
# This code is contributed by mohit kumar 29
C
// C# program for the above approach
using System;
class GFG{
// Function to check if the given
// string is linear or not
static bool is_linear(String s)
{
int tmp = 0;
char first = s[0];
// Iterate over string
for(int pos = 0; pos < s.Length;
pos += tmp)
{
// If character is not same as
// the first character then
// return false
if (s[pos] != first)
{
return false;
}
// Increment the tmp
tmp++;
}
return true;
}
// Driver code
public static void Main(String[] args)
{
// Given String str
String str = "aapaxyayziabcde";
// Function call
if (is_linear(str))
{
Console.Write("Yes");
}
else
{
Console.Write("No");
}
}
}
// This code is contributed by grand_master
java 描述语言
<script>
// Javascript program for the above approach
// Function to check if the given
// string is linear or not
function is_linear(s)
{
let tmp = 0;
let first = s[0];
// Iterate over string
for(let pos = 0; pos < s.length;
pos += tmp)
{
// If character is not same as
// the first character then
// return false
if (s[pos] != first)
{
return false;
}
// Increment the tmp
tmp++;
}
return true;
}
// Driver Code
// Given String str
let str = "aapaxyayziabcde";
// Function call
if (is_linear(str))
{
document.write("Yes");
}
else
{
document.write("No");
}
</script>
Output:
Yes
时间复杂度:O(log N) T5】辅助空间: O(1)
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