以“极客”开头、以“for”结尾的子字符串数量
原文:https://www . geeksforgeeks . org/以 geeks 开头和以 for 结尾的子字符串数量/
给定一个由小写英文字母组成的字符串 str ，任务是找出以“极客”开头，以“表示”结尾的子字符串的数量。 举例:
输入:str = " geeksforgeicks " 输出:3 “geeks for”“geeksforgeicks for”“geeks for” 是唯一有效的子字符串。 输入:str = " geeks forgeeks " 输出: 1
天真的方法:首先将计数器设置为 0，然后遍历字符串，每当遇到子字符串“极客”时，从下一个索引再次遍历字符串，并尝试找到“”的子字符串。如果“for”存在，则递增计数器并最终打印。 高效方法:为子串“极客”和“为”设置两个计数器，比如 c1 和 c2。迭代时，每当遇到子串“geeks”时，递增 c1，每当遇到“for”时，设置 c2 = c2 + c1 。这是因为“极客”的每一次出现都会用当前找到的做一个有效的子串来表示。最后打印 c2 。 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include <iostream>
using namespace std;
// Function to return the count
// of required substrings
int countSubStr(string s, int n)
{
int c1 = 0, c2 = 0;
// For every index of the string
for (int i = 0; i < n; i++) {
// If the substring starting at
// the current index is "geeks"
if (s.substr(i, 5) == "geeks")
c1++;
// If the substring is "for"
if (s.substr(i, 3) == "for")
c2 = c2 + c1;
}
return c2;
}
// Driver code
int main()
{
string s = "geeksforgeeksisforgeeks";
int n = s.size();
cout << countSubStr(s, n);
return 0;
}
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 to return the count
// of required substrings
static int countSubStr(String s, int n)
{
int c1 = 0, c2 = 0;
// For every index of the string
for (int i = 0; i < n; i++)
{
// If the substring starting at
// the current index is "geeks"
if (i < n - 5 &&
"geeks".equals(s.substring(i, i + 5)))
{
c1++;
}
// If the substring is "for"
if (i < n - 3 &&
"for".equals(s.substring(i, i + 3)))
{
c2 = c2 + c1;
}
}
return c2;
}
// Driver code
public static void main(String[] args)
{
String s = "geeksforgeeksisforgeeks";
int n = s.length();
System.out.println(countSubStr(s, n));
}
}
// This code is contributed by 29AjayKumar
Python 3
# Python3 implementation of the approach
# Function to return the count
# of required substrings
def countSubStr(s, n) :
c1 = 0; c2 = 0;
# For every index of the string
for i in range(n) :
# If the substring starting at
# the current index is "geeks"
if (s[i : i + 5] == "geeks") :
c1 += 1;
# If the substring is "for"
if (s[i :i+ 3] == "for") :
c2 = c2 + c1;
return c2;
# Driver code
if __name__ == "__main__" :
s = "geeksforgeeksisforgeeks";
n = len(s);
print(countSubStr(s, n));
# This code is contributed by AnkitRai01
C
// C# implementation of the approach
using System;
public class GFG
{
// Function to return the count
// of required substrings
static int countSubStr(String s, int n)
{
int c1 = 0, c2 = 0;
// For every index of the string
for (int i = 0; i < n; i++)
{
// If the substring starting at
// the current index is "geeks"
if (i < n - 5 &&
"geeks".Equals(s.Substring(i, 5)))
{
c1++;
}
// If the substring is "for"
if (i < n - 3 &&
"for".Equals(s.Substring(i, 3)))
{
c2 = c2 + c1;
}
}
return c2;
}
// Driver code
public static void Main(String[] args)
{
String s = "geeksforgeeksisforgeeks";
int n = s.Length;
Console.WriteLine(countSubStr(s, n));
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// JavaScript implementation of the approach
// Function to return the count
// of required substrings
function countSubStr(s, n)
{
var c1 = 0, c2 = 0;
// For every index of the string
for (var i = 0; i < n; i++) {
// If the substring starting at
// the current index is "geeks"
if (s.substring(i, i+5) == "geeks")
c1++;
// If the substring is "for"
if (s.substring(i,i+ 3) == "for")
c2 = c2 + c1;
}
return c2;
}
// Driver code
var s = "geeksforgeeksisforgeeks";
var n = s.length;
document.write( countSubStr(s, n));
</script>
Output:
3
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