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📜  找到具有最大乘积的子串

📅  最后修改于: 2021-09-03 13:44:43             🧑  作者: Mango

给定字符串str仅包含大小为N 的小写英文字母,任务是找到具有最大乘积的子字符串。

例子:

方法:

  • 首先,遍历给定的字符串,同时保持最大的乘积值。
  • 乘积将始终保持增加或保持不变,除非我们遇到‘a’ 。因此,在每次出现‘a’后开始一个新的子字符串。
  • 此外,除了最大乘积值,我们还将维护最大乘积对应的子字符串。
  • 遍历整个字符串,打印与最大乘积对应的子字符串。

下面是上述方法的实现:

C++
// C++ program to find the maximum product substring
 
#include 
using namespace std;
 
// Function to return the value of a character
int value(char x)
{
    return (int)(x - 'a');
}
 
// Function to find the maximum product substring
string maximumProduct(string str, int n)
{
    // To store substrings
    string answer = "", curr = "";
    long long maxProduct = 0, product = 1;
 
    for (int i = 0; i < n; i++) {
        product *= 1LL * value(str[i]);
 
        curr += str[i];
 
        // Check if current product is maximum
        // possible or not
        if (product >= maxProduct) {
            maxProduct = product;
            answer = curr;
        }
 
        // If product is 0
        if (product == 0) {
            product = 1;
            curr = "";
        }
    }
 
    // Return the substring with maximum product
    return answer;
}
 
// Driver code
int main()
{
    string str = "sdtfakdhdahdzz";
 
    int n = str.size();
 
    // Function call
    cout << maximumProduct(str, n) << endl;
 
    return 0;
}


Java
// Java program to find the maximum product subString
 
class GFG{
  
// Function to return the value of a character
static int value(char x)
{
    return (int)(x - 'a');
}
  
// Function to find the maximum product subString
static String maximumProduct(String str, int n)
{
    // To store subStrings
    String answer = "", curr = "";
    long maxProduct = 0, product = 1;
  
    for (int i = 0; i < n; i++) {
        product *= 1L * value(str.charAt(i));
  
        curr += str.charAt(i);
  
        // Check if current product is maximum
        // possible or not
        if (product >= maxProduct) {
            maxProduct = product;
            answer = curr;
        }
  
        // If product is 0
        if (product == 0) {
            product = 1;
            curr = "";
        }
    }
  
    // Return the subString with maximum product
    return answer;
}
  
// Driver code
public static void main(String[] args)
{
    String str = "sdtfakdhdahdzz";
  
    int n = str.length();
  
    // Function call
    System.out.print(maximumProduct(str, n) +"\n");
  
}
}
 
// This code is contributed by 29AjayKumar


Python3
# Python3 program to find the maximum product sub
 
# Function to return the value of a character
def value(x):
    return (ord(x) - ord('a'))
 
# Function to find the maximum product sub
def maximumProduct(strr, n):
 
    # To store subs
    answer = ""
    curr = ""
    maxProduct = 0
    product = 1
 
    for i in range(n):
        product *=value(strr[i])
 
        curr += strr[i]
 
        # Check if current product is maximum
        # possible or not
        if (product >= maxProduct):
            maxProduct = product
            answer = curr
 
        # If product is 0
        if (product == 0):
            product = 1
            curr = ""
 
    # Return the sub with maximum product
    return answer
 
# Driver code
if __name__ == '__main__':
    strr = "sdtfakdhdahdzz"
 
    n = len(strr)
 
    # Function call
    print(maximumProduct(strr, n))
 
# This code is contributed by mohit kumar 29


C#
// C# program to find the maximum product subString
 
using System;
 
public class GFG{
 
// Function to return the value of a character
static int value(char x)
{
    return (int)(x - 'a');
}
 
// Function to find the maximum product subString
static String maximumProduct(String str, int n)
{
    // To store subStrings
    String answer = "", curr = "";
    long maxProduct = 0, product = 1;
 
    for (int i = 0; i < n; i++) {
        product *= 1L * value(str[i]);
 
        curr += str[i];
 
        // Check if current product is maximum
        // possible or not
        if (product >= maxProduct) {
            maxProduct = product;
            answer = curr;
        }
 
        // If product is 0
        if (product == 0) {
            product = 1;
            curr = "";
        }
    }
 
    // Return the subString with maximum product
    return answer;
}
 
// Driver code
public static void Main(String[] args)
{
    String str = "sdtfakdhdahdzz";
 
    int n = str.Length;
 
    // Function call
    Console.Write(maximumProduct(str, n) +"\n");
 
}
}
// This code is contributed by PrinciRaj1992


Javascript


输出:
hdzz

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