使用字符字符的最长公共前缀
给定一组字符串，找出最长的公共前缀。
Input : {“geeksforgeeks”, “geeks”, “geek”, “geezer”}
Output : "gee"
Input : {"apple", "ape", "april"}
Output : "ap"我们在上一篇文章中讨论了逐字匹配算法。
在这个算法中，我们不是一个一个地遍历字符串，而是一个一个地遍历字符。
我们将我们的字符串视为 – “geeksforgeeks”、“geeks”、“geek”、“geezer”。
下面是这种方法的实现。
C++
// A C++ Program to find the longest common prefix
#include
using namespace std;
// A Function to find the string having the minimum
// length and returns that length
int findMinLength(string arr[], int n)
{
int min = arr[0].length();
for (int i=1; i Java
// A Java Program to find the longest common prefix
class GFG
{
// A Function to find the string having the minimum
// length and returns that length
static int findMinLength(String arr[], int n)
{
int min = arr[0].length();
for (int i = 1; i < n; i++)
{
if (arr[i].length() < min)
{
min = arr[i].length();
}
}
return (min);
}
// A Function that returns the longest common prefix
// from the array of strings
static String commonPrefix(String arr[], int n)
{
int minlen = findMinLength(arr, n);
String result = ""; // Our resultant string
char current; // The current character
for (int i = 0; i < minlen; i++)
{
// Current character (must be same
// in all strings to be a part of
// result)
current = arr[0].charAt(i);
for (int j = 1; j < n; j++)
{
if (arr[j].charAt(i) != current)
{
return result;
}
}
// Append to result
result += (current);
}
return (result);
}
// Driver program to test above function
public static void main(String[] args)
{
String arr[] = {"geeksforgeeks", "geeks",
"geek", "geezer"};
int n = arr.length;
String ans = commonPrefix(arr, n);
if (ans.length() > 0) {
System.out.println("The longest common prefix is "
+ ans);
} else {
System.out.println("There is no common prefix");
}
}
}
// This code contributed by Rajput-JiPython 3
# Python 3 Program to find the longest common prefix
# A Function to find the string having the minimum
# length and returns that length
def findMinLength(arr, n):
min = len(arr[0])
for i in range(1,n):
if (len(arr[i])< min):
min = len(arr[i])
return(min)
# A Function that returns the longest common prefix
# from the array of strings
def commonPrefix(arr, n):
minlen = findMinLength(arr, n)
result =""
for i in range(minlen):
# Current character (must be same
# in all strings to be a part of
# result)
current = arr[0][i]
for j in range(1,n):
if (arr[j][i] != current):
return result
# Append to result
result = result+current
return (result)
# Driver program to test above function
if __name__ == "__main__":
arr = ["geeksforgeeks", "geeks",
"geek", "geezer"]
n = len(arr)
ans = commonPrefix (arr, n)
if (len(ans)):
print("The longest common prefix is ",ans)
else:
print("There is no common prefix")C#
// A C# Program to find the longest common prefix
using System;
class GFG
{
// A Function to find the string having the minimum
// length and returns that length
static int findMinLength(String []arr, int n)
{
int min = arr[0].Length;
for (int i = 1; i < n; i++)
{
if (arr[i].Length < min)
{
min = arr[i].Length;
}
}
return (min);
}
// A Function that returns the longest common prefix
// from the array of strings
static String commonPrefix(String []arr, int n)
{
int minlen = findMinLength(arr, n);
String result = ""; // Our resultant string
char current; // The current character
for (int i = 0; i < minlen; i++)
{
// Current character (must be same
// in all strings to be a part of
// result)
current = arr[0][i];
for (int j = 1; j < n; j++)
{
if (arr[j][i] != current)
{
return result;
}
}
// Append to result
result += (current);
}
return (result);
}
// Driver code
public static void Main(String[] args)
{
String []arr = {"geeksforgeeks", "geeks",
"geek", "geezer"};
int n = arr.Length;
String ans = commonPrefix(arr, n);
if (ans.Length > 0)
{
Console.WriteLine("The longest common prefix is "
+ ans);
}
else
{
Console.WriteLine("There is no common prefix");
}
}
}
/* This code contributed by PrinciRaj1992 */Javascript
输出 ：
The longest common prefix is gee这个算法比“逐字匹配”算法好多少？-
在第 1 组中，我们讨论了“逐字匹配”算法。
假设您的输入字符串为“geeksforgeeks”、“geeks”、“geek”、“geezer”、“x”。
现在没有上述字符串的公共前缀字符串。通过 Set 1 中讨论的“逐字匹配”算法，我们通过遍历所有字符串得出没有公共前缀字符串的结论。但是如果我们使用这个算法，那么在第一次迭代中我们就会知道没有公共前缀字符串 ，因为我们不会进一步寻找每个字符串的第二个字符。
当字符串太多时，这种算法有很大的优势。
时间复杂度：由于我们正在遍历所有字符的所有字符串，所以我们可以说时间复杂度是 O(NM) 其中，
N = Number of strings
M = Length of the largest string string