组移位字符串

给定一个字符串数组（全部为小写字母），任务是以这样一种方式将它们分组，即一组中的所有字符串都是彼此的移位版本。如果两个字符串S 和 T 被称为移位，

S.length = T.length 
and
S[i] = T[i] + K for 
1 <= i <= S.length  for a constant integer K

例如，字符串、{acd、dfg、wyz、yab、mop} 是彼此的移位版本。

Input  : str[] = {"acd", "dfg", "wyz", "yab", "mop",
                 "bdfh", "a", "x", "moqs"};

Output : a x
         acd dfg wyz yab mop
         bdfh moqs
All shifted strings are grouped together.

我们可以看到一组的字符串中的模式，该字符串的所有的字符连续字符之间的差别是相等的。如上例所示，取 acd、dfg 和 mop
acd -> 2 1
dfg -> 2 1
拖把 -> 2 1
由于差异相同，我们可以使用它来识别属于同一组的字符串。这个想法是形成字符串差异作为关键。如果找到具有相同的差异字符串的字符串，那么这个字符串也属于同一组。例如，上面三个字符串有相同的差异字符串，即“21”。
在下面的实现中，我们将 ‘a’ 添加到每个差异并将 21 存储为“ba”。

C++

/* C/C++ program to print groups of shifted strings
   together. */
#include 
using namespace std;
const int ALPHA = 26;   // Total lowercase letter
 
// Method to a difference string for a given string.
// If string is "adf" then difference string will be
// "cb" (first difference 3 then difference 2)
string getDiffString(string str)
{
    string shift = "";
    for (int i = 1; i < str.length(); i++)
    {
        int dif = str[i] - str[i-1];
        if (dif < 0)
            dif += ALPHA;
 
        // Representing the difference as char
        shift += (dif + 'a');
    }
 
    // This string will be 1 less length than str
    return shift;
}
 
// Method for grouping shifted string
void groupShiftedString(string str[], int n)
{
    // map for storing indices of string which are
    // in same group
    map< string, vector > groupMap;
    for (int i = 0; i < n; i++)
    {
        string diffStr = getDiffString(str[i]);
        groupMap[diffStr].push_back(i);
    }
 
    // iterating through map to print group
    for (auto it=groupMap.begin(); it!=groupMap.end();
                                                it++)
    {
        vector v = it->second;
        for (int i = 0; i < v.size(); i++)
            cout << str[v[i]] << " ";
        cout << endl;
    }
}
 
// Driver method to test above methods
int main()
{
    string str[] = {"acd", "dfg", "wyz", "yab", "mop",
                    "bdfh", "a", "x", "moqs"
                   };
    int n = sizeof(str)/sizeof(str[0]);
    groupShiftedString(str, n);
    return 0;
}

Python3

# Python3 program to print groups
# of shifted strings together.
 
# Total lowercase letter
ALPHA = 26
 
# Method to a difference string
# for a given string. If string
# is "adf" then difference string
# will be "cb" (first difference
# 3 then difference 2)
def getDiffString(str):
   
    shift=""
 
    for i in range(1, len(str)):
        dif = (ord(str[i]) -
              ord(str[i - 1]))
 
        if(dif < 0):
            dif += ALPHA
 
        # Representing the difference
        # as char
        shift += chr(dif + ord('a'))
 
    # This string will be 1 less
    # length than str
    return shift
 
# Method for grouping
# shifted string
def groupShiftedString(str,n):
 
    # map for storing indices
    # of string which are
    # in same group
    groupMap = {}
 
    for i in range(n):
        diffStr = getDiffString(str[i])
        if diffStr not in groupMap:
            groupMap[diffStr] = [i]
        else:
            groupMap[diffStr].append(i)
     
    # Iterating through map
    # to print group
    for it in groupMap:
        v = groupMap[it]
        for i in range(len(v)):
            print(str[v[i]], end = " ")
        print()
         
# Driver code
str = ["acd", "dfg", "wyz",
       "yab", "mop","bdfh",
       "a", "x", "moqs"]
n = len(str)
groupShiftedString(str, n)
 
# This code is contributed by avanitrachhadiya2155

Javascript

 
/* Javascript program to print groups of shifted strings
   together. */
let ALPHA = 26;
 
// Method to a difference string for a given string.
// If string is "adf" then difference string will be
// "cb" (first difference 3 then difference 2)
function getDiffString(str)
{
    let shift = "";
    for (let i = 1; i < str.length; i++)
    {
        let dif = str[i] - str[i-1];
        if (dif < 0)
            dif += ALPHA;
  
        // Representing the difference as char
        shift += (dif + 'a');
    }
  
    // This string will be 1 less length than str
    return shift;
}
 
// Method for grouping shifted string
function groupShiftedString(str, n)
{
 
    // map for storing indices of string which are
    // in same group
    let groupMap = new Map();
    for (let i = 0; i < n; i++)
    {
        let diffStr = getDiffString(str[i]);
        if(!groupMap.has(diffStr))
            groupMap.set(diffStr,[]);   
        groupMap.get(diffStr).push(i);
    }
  
    // iterating through map to print group
    for (let [key, value] of groupMap.entries())
    {
        let v = value;
        for (let i = 0; i < v.length; i++)
            document.write(str[v[i]]+" ");
        document.write("<br>")
    }
}
 
// Driver method to test above methods
let str=["acd", "dfg", "wyz", "yab", "mop",
                    "bdfh", "a", "x", "moqs"];
let n = str.length;
groupShiftedString(str, n);
 
// This code is contributed by ab2127

输出：

a x
acd dfg wyz yab mop
bdfh moqs

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