在给定字符串str的情况下，任务是对给定字符串的所有双音子字符串进行计数。
A bitonic substring is a substring of the given string in which elements are either strictly increasing or strictly decreasing, or first increasing and then decreasing.
例子：
Input: str = “bade”
Output: 8
Explanation:
Substrings of length 1 are always bitonic, “b”, “a”, “d”, “e”
Substrings of length 2 are “ba”, “ad”, “de” and these all are also bitonic because they are increasing or decresing only.
Substrings of length 3 are “bad “and “ade” in which “ade” is bitonic.
Substring of length 4 “bade” is not bitonic because it decreases and then increases.
So total 8 substrings are bitonic.
Input: str = “abc”
Output: 6
Explanation:
The given string is increasing, so all it’s substrings are also increasing and hence are bitonic. So total = 6.
方法：想法是生成给定字符串的所有可能的子字符串，并检查每个子字符串是否为Bitonic。如果子字符串是Bitonic，则增加答案的计数。最后，打印所有双子位子阵列的计数。
下面是上述方法的实现：
C++
// C+++ program for the above approach
#include
using namespace std;
// Function to find all the bitonic
// sub strings
void subString(char str[], int n)
{
// Pick starting point
int c = 0;
// Iterate till length of the string
for (int len = 1; len <= n; len++) {
// Pick ending point for string
for (int i = 0; i <= n - len; i++) {
// Substring from i to j
// is obtained
int j = i + len - 1;
char temp = str[i], f = 0;
// Substrings of length 1
if (j == i) {
// Increase count
c++;
continue;
}
int k = i + 1;
// For increasing sequence
while (temp < str[k] && k <= j) {
temp = str[k];
k++;
f = 2;
}
// Check for strictly increasing
if (k > j) {
// Increase count
c++;
f = 2;
}
// Check for decreasing sequence
while (temp > str[k]
&& k <= j
&& f != 2) {
k++;
f = 0;
}
if (k > j && f != 2) {
// Increase count
c++;
f = 0;
}
}
}
// Print the result
cout << c << endl;
}
// Driver Code
int main()
{
// Given string
char str[] = "bade";
// Function Call
subString(str, strlen(str));
return 0;
} Java
// Java+ program for the above approach
import java.util.*;
class GFG{
// Function to find all the bitonic
// sub Strings
static void subString(char str[], int n)
{
// Pick starting point
int c = 0;
// Iterate till length of the String
for(int len = 1; len <= n; len++)
{
// Pick ending point for String
for(int i = 0; i <= n - len; i++)
{
// SubString from i to j
// is obtained
int j = i + len - 1;
char temp = str[i], f = 0;
// SubStrings of length 1
if (j == i)
{
// Increase count
c++;
continue;
}
int k = i + 1;
// For increasing sequence
while (k < n && temp < str[k])
{
temp = str[k];
k++;
f = 2;
}
// Check for strictly increasing
if (k > j)
{
// Increase count
c++;
f = 2;
}
// Check for decreasing sequence
while (k < n && temp > str[k] &&
f != 2)
{
k++;
f = 0;
}
if (k > j && f != 2)
{
// Increase count
c++;
f = 0;
}
}
}
// Print the result
System.out.print(c + "\n");
}
// Driver Code
public static void main(String[] args)
{
// Given String
char str[] = "bade".toCharArray();
// Function call
subString(str, str.length);
}
}
// This code is contributed by 29AjayKumarPython3
# Python3 program for the above approach
# Function to find all the bitonic
# sub strings
def subString(str, n):
# Pick starting po
c = 0;
# Iterate till length of the string
for len in range(1, n + 1):
# Pick ending pofor string
for i in range(0, n - len + 1):
# Substring from i to j
# is obtained
j = i + len - 1;
temp = str[i]
f = 0;
# Substrings of length 1
if (j == i):
# Increase count
c += 1
continue;
k = i + 1;
# For increasing sequence
while (k <= j and temp < str[k]):
temp = str[k];
k += 1;
f = 2;
# Check for strictly increasing
if (k > j):
# Increase count
c += 1;
f = 2;
# Check for decreasing sequence
while (k <= j and temp > str[k] and
f != 2):
k += 1;
f = 0;
if (k > j and f != 2):
# Increase count
c += 1;
f = 0;
# Print the result
print(c)
# Driver code
# Given string
str = "bade";
# Function Call
subString(str, len(str))
# This code is contributed by grand_masterC#
// C# program for the above approach
using System;
class GFG{
// Function to find all the bitonic
// sub Strings
static void subString(char []str, int n)
{
// Pick starting point
int c = 0;
// Iterate till length of the String
for(int len = 1; len <= n; len++)
{
// Pick ending point for String
for(int i = 0; i <= n - len; i++)
{
// SubString from i to j
// is obtained
int j = i + len - 1;
char temp = str[i], f = (char)0;
// SubStrings of length 1
if (j == i)
{
// Increase count
c++;
continue;
}
int k = i + 1;
// For increasing sequence
while (k < n && temp < str[k])
{
temp = str[k];
k++;
f = (char)2;
}
// Check for strictly increasing
if (k > j)
{
// Increase count
c++;
f = (char)2;
}
// Check for decreasing sequence
while (k < n && temp > str[k] &&
f != 2)
{
k++;
f = (char)0;
}
if (k > j && f != 2)
{
// Increase count
c++;
f = (char)0;
}
}
}
// Print the result
Console.Write(c + "\n");
}
// Driver Code
public static void Main(String[] args)
{
// Given String
char []str = "bade".ToCharArray();
// Function call
subString(str, str.Length);
}
}
// This code is contributed by 29AjayKumar输出：
8
时间复杂度： O(N 2 )
辅助空间： O(1)