统计数值大于 X 的子串数量
原文:https://www . geesforgeks . org/count-number-of-substrings-with-numeric-value-behind-x/
给定一个字符串“S”(由数字组成)和一个整数“X”，任务是计算满足以下条件的“S”的所有子字符串:
- 子字符串不能以数字“0”开头。
- 它所代表的数字必须大于“X”。
注意:如果两种选择子串的方式在不同的索引处开始或结束，则它们是不同的。
示例:
Input: S = "471", X = 47
Output: 2
Only the sub-strings "471" and "71"
satisfy the given conditions.
Input: S = "2222", X = 97
Output: 3
Valid strings are "222", "222" and "2222".
进场:
- 迭代字符串“S”的每个数字，并选择大于“0”的数字。
- 现在，从上一步中选择的字符开始获取所有可能的子字符串，并将每个子字符串转换为整数。
- 将上一步中的整数与“X”进行比较。如果数字大于“X”，则递增计数变量。
- 最后，打印计数变量的值。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function that counts
// valid sub-strings
int count(string S, int X)
{
int count = 0;
const int N = S.length();
for (int i = 0; i < N; ++i) {
// Only take those numbers
// that do not start with '0'.
if (S[i] != '0') {
for (int len = 1; (i + len) <= N; ++len) {
// converting the sub-string
// starting from index 'i'
// and having length 'len' to int
// and checking if it is greater
// than X or not
if (stoi(S.substr(i, len)) > X)
count++;
}
}
}
return count;
}
// Driver code
int main()
{
string S = "2222";
int X = 97;
cout << count(S, X);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function that counts
// valid sub-strings
static int count(String S, int X)
{
int count = 0;
int N = S.length();
for (int i = 0; i < N; ++i)
{
// Only take those numbers
// that do not start with '0'.
if (S.charAt(i) != '0')
{
for (int len = 1;
(i + len) <= N; ++len)
{
// converting the sub-string
// starting from index 'i'
// and having length 'len' to
// int and checking if it is
// greater than X or not
int num = Integer.parseInt(S.substring(i, i + len));
if (num > X)
count++;
}
}
}
return count;
}
// Driver code
public static void main(String []args)
{
String S = "2222";
int X = 97;
System.out.println(count(S, X));
}
}
// This code is contributed by ihritik
Python 3
# Python3 implementation of
# the approach
# Function that counts
# valid sub-strings
def countSubStr(S, X):
cnt = 0
N = len(S)
for i in range(0, N):
# Only take those numbers
# that do not start with '0'.
if (S[i] != '0'):
j = 1
while((j + i) <= N):
# converting the sub-string
# starting from index 'i'
# and having length 'len' to
# int and checking if it is
# greater than X or not
num = int(S[i : i + j])
if (num > X):
cnt = cnt + 1
j = j + 1
return cnt;
# Driver code
S = "2222";
X = 97;
print(countSubStr(S, X))
# This code is contributed by ihritik
C
// C# implementation of the approach
using System;
class GFG
{
// Function that counts
// valid sub-strings
static int count(string S, int X)
{
int count = 0;
int N = S.Length;
for (int i = 0; i < N; ++i)
{
// Only take those numbers
// that do not start with '0'.
if (S[i] != '0')
{
for (int len = 1;
(i + len) <= N; ++len)
{
// converting the sub-string
// starting from index 'i'
// and having length 'len' to int
// and checking if it is greater
// than X or not
int num = Int32.Parse(S.Substring(i, len));
if (num > X)
count++;
}
}
}
return count;
}
// Driver code
public static void Main(String []args)
{
string S = "2222";
int X = 97;
Console.WriteLine(count(S, X));
}
}
// This code is contributed by ihritik
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function that counts
// valid sub-strings
function countSubStr($S, $X)
{
$cnt = 0;
$N = strlen($S);
for ($i = 0; $i < $N; ++$i)
{
// Only take those numbers
// that do not start w$ith '0'.
if ($S[$i] != '0')
{
for ($len = 1;
($i + $len) <= $N; ++$len)
{
// converting the sub-str$ing
// starting from index 'i'
// and having length 'len' to int
// and checking if it is greater
// than X or not
$num = intval(substr($S, $i, $len));
if ($num > $X)
$cnt++;
}
}
}
return $cnt;
}
// Driver code
$S = "2222";
$X = 97;
echo countSubStr($S, $X);
// This code is contributed by ihritik
?>
java 描述语言
<script>
// Javascript implementation of the approach
// Function that counts
// valid sub-strings
function count(S, X)
{
var count = 0;
var N = S.length;
for (var i = 0; i < N; ++i) {
// Only take those numbers
// that do not start with '0'.
if (S[i] != '0') {
for (var len = 1; (i + len) <= N; ++len) {
// converting the sub-string
// starting from index 'i'
// and having length 'len' to int
// and checking if it is greater
// than X or not
if (parseInt(S.substring(i, i+len)) > X)
count++;
}
}
}
return count;
}
// Driver code
var S = "2222";
var X = 97;
document.write( count(S, X));
</script>
Output:
3
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