最长公共前缀_最长的公共前缀

最长公共前缀

Problem statement:

问题陈述：

Write a function to find the longest common prefix string amongst an array of strings.

编写函数以在字符串数组中找到最长的公共前缀字符串 。

If there is no common prefix, return an empty string "".

如果没有公共前缀，则返回一个空字符串“” 。

Solution:

解：

Input Example:

输入示例：

    Example 1:
Let the set of strings to be {"flower", "fly", "flow"}
The longest common prefix is "fl"
Example 2:
Let the set of strings to be {"dog", "donkey", "door", "cat"}
The longest common prefix is "", i.e., empty string. 
Since there is no common prefix at all.

最长的公共前缀 (Longest common prefix)

Longest common prefix simply means the longest prefix (prefix is a substring also, but not vice-versa) all the member strings consist of.

最长公共前缀仅表示所有成员字符串组成的最长前缀(前缀也是子字符串，反之亦然)。

查找一组字符串的最长公共前缀的算法 (Algorithm to find longest common prefix of a set of strings)

Solving particularly for two string, the problem is not that difficult what it is for a set of strings. But we can simply break the problem into sub-problems where we can solve for only two strings and simply pass the output for further string processing 's. In a simple word, the whole thing can be formulated to be:

特别是对于两个字符串，解决的问题并不像一组字符串那么困难。但是我们可以简单地将问题分解为几个子问题，在这些子问题中，我们只可以解决两个字符串，然后将输出传递给进一步的字符串处理。简而言之，整个事情可以表述为：

    LongestCommonPrefix(string1, string2, ..., string n)
=   LongestCommonPrefix(LongestCommonPrefix(string1,string2),string3,...,string n)
=   LongestCommonPrefix(newstring1,string3,string4,...,string n)
=   ...
=   LongestCommonPrefix(newstring n-1, string n)
=   new string n = final result

So this simply converts the problem for a set of 
strings to a problem of two strings only

Finding longest common prefix for two strings (string x, string y)

查找两个字符串(字符串x，字符串y)的最长公共前缀

Check for base case
检查基本情况

IF anyone is empty string
return empty string;

Declare result as an empty string;
将结果声明为空字符串；

    IF string x is smaller than y
//check for common prefix part on basis of x
FOR( i=0;i<length of string x; i++)
IF(x[i]==y[i])
result+=x[i]; //append the common character
ELSE//no common character at this point
Returnresult
END FOR
ELSE//if string y is smaller than x
//check for common prefix part on basis of y		
FOR (i=0;i<length of y; i++)
IF(y[i]==x[i])
result+=y[i];//append the common character
ELSE//no common character at this point
return result;
END FOR
END IF-ELSE

result consist of longest common prefix for these two strings
结果由这两个字符串的最长公共前缀组成

Explanation with example

举例说明

    Let's consider the first example
The set of strings: {"flower", "fly", "flow"}
LongestCommonPrefix("flower", "fly", "flow")
=   LongestCommonPrefix(LongestCommonPrefix ("flower","fly"),"flow")
=   LongestCommonPrefix("fl", "flow")
=   "fl"
Let's consider the second example
the set of strings to be {"dog", "donkey", "door", "cat"}
LongestCommonPrefix("dog", "donkey", "door", "cat")
=   LongestCommonPrefix(LongestCommonPrefix ("dog", "donkey"),"door", "cat")
=   LongestCommonPrefix("do","door", "cat")
=   LongestCommonPrefix (LongestCommonPrefix("do", "door") , "cat")
=   LongestCommonPrefix("do", "cat")
=   "" //empty string

C++ implementation

C ++实现

#include <bits/stdc++.h>
using namespace std;
string findPrefix(string x,string y){
//base case checking
if(x=="" || y=="")
return "";
string result="";
//if string x is smaller than y
if(x.length()<y.length()){
//check up for common prefix part
for(int i=0;i<x.length();i++){
if(x[i]==y[i])
result+=x[i];
}
}
else{
//if string y is smaller than x
for(int i=0;i<y.length();i++){
//check up for common prefix part 
if(y[i]==x[i])
result+=y[i];
else
return result;
}
}
return result;
}
string longestCommonPrefix(vector<string>& strs) {
//base cases
if(strs.size()==0)
return "";
//if only one string exists
if(strs.size()==1)
return strs[0];
string prefix=strs[0];
//follow the associative property for checking 
//take two strings each time & pass output prefix 
//as one string for further processings
for(int i=1;i<strs.size();i++){
prefix=findPrefix(prefix,strs[i]);
if(prefix=="")
return prefix;
}
return prefix;
}
int main(){
int n;
cout<<"enter no of strings you want to add\n";
cin>>n;
string s;
vector<string> v;
cout<<"enter "<<n<<" strings\n";
//collect input
while(n--){
cin>>s;
v.push_back(s);
}
//print longest common prefix
if(longestCommonPrefix(v)=="")
cout<<"no common prefix between the strings\n";
else
cout<<"longest common prefix: "<<longestCommonPrefix(v)<<endl;
return 0;
}

Output

输出量

First run:
enter no of strings you want to add
3
enter 3 strings
flower
fly
flow
longest common prefix: fl
Second run:
enter no of strings you want to add
4
enter 4 strings
dog
donkey
door
cat
no common prefix between the strings