给定一个字符串S，任务是找到的最小长度字符串，该字符串中相邻字符保持在最小长度的字符串相邻。
例子：
Input: S = “acabpba”
Output: pbac
Explanation:
The given string can be converted to “pbac” in which,
every adjacent character remains adjacent.
Input: S = “abcdea”
Output: Impossible
Explanation:
It is impossible to find such string.
方法：思想是准备一个类似图的结构，其中图的每个相邻节点表示字符串的相邻字符。在两种情况下，这种类型的字符串是不可能的——
- 如果一个字符包含三个或更多相邻字符。
- 如果两个字符不只有一个相邻字符，长度为 1 的字符串除外。
如果字符串的上述条件为真，则只需使用深度优先搜索遍历遍历图形，此遍历的路径将是最小长度的字符串。为DFS源顶点将是字符仅具有一个相邻的字符的任何一个。
下面是上述方法的实现：
C++
// C++ implementation to find the
// minimum length string such that
// adjacent characters of the string
// remains adjacent in the string
#include
using namespace std;
class graph {
int arr[26][2];
vector alpha;
vector answer;
public:
// Constructor for the graph
graph()
{
// Initialize the matrix by -1
for (int i = 0; i < 26; i++) {
for (int j = 0; j < 2; j++)
arr[i][j] = -1;
}
// Initialize the alphabet array
alpha = vector(26);
}
// Function to do Depth first
// search Traversal
void dfs(int v)
{
// Pushing current character
answer.push_back(v);
alpha[v] = 1;
for (int i = 0; i < 2; i++) {
if (alpha[arr[v][i]] == 1
|| arr[v][i] == -1)
continue;
dfs(arr[v][i]);
}
}
// Function to find the minimum
// length string
void minString(string str)
{
// Condition if given string
// length is 1
if (str.length() == 1) {
cout << str << "\n";
return;
}
bool flag = true;
// Loop to find the adjacency
// list of the given string
for (int i = 0; i < str.length() - 1;
i++) {
int j = str[i] - 'a';
int k = str[i + 1] - 'a';
// Condition if character
// already present
if (arr[j][0] == k
|| arr[j][1] == k) {
}
else if (arr[j][0] == -1)
arr[j][0] = k;
else if (arr[j][1] == -1)
arr[j][1] = k;
// Condition if a character
// have more than two different
// adjacent characters
else {
flag = false;
break;
}
if (arr[k][0] == j
|| arr[k][1] == j) {
}
else if (arr[k][0] == -1)
arr[k][0] = j;
else if (arr[k][1] == -1)
arr[k][1] = j;
// Condition if a character
// have more than two different
// adjacent characters
else {
flag = false;
break;
}
}
// Variable to check string contain
// two end characters or not
bool contain_ends = false;
int count = 0;
int index;
for (int i = 0; i < 26; i++) {
// Condition if a character has
// only one type of adjacent
// character
if ((arr[i][0] == -1
&& arr[i][1] != -1)
|| (arr[i][1] == -1
&& arr[i][0] != -1)) {
count++;
index = i;
}
// Condition if the given string
// has exactly two characters
// having only one type of adjacent
// character
if (count == 2)
contain_ends = true;
if (count == 3)
contain_ends = false;
}
if (contain_ends == false
|| flag == false) {
cout << "Impossible"
<< "\n";
return;
}
// Depth first Search Traversal
// on one of the possible end
// character of the string
dfs(index);
// Loop to print the answer
for (int i = 0; i < answer.size();
i++) {
char ch = answer[i] + 'a';
cout << ch;
}
}
};
// Driver Code
int main()
{
string s = "acabpba";
graph g;
g.minString(s);
return 0;
} Java
// Java implementation to find the
// minimum length string such that
// adjacent characters of the string
// remains adjacent in the string
import java.util.ArrayList;
class Graph{
int[][] arr;
ArrayList alpha;
ArrayList answer;
// Constructor for the Graph
public Graph()
{
arr = new int[26][2];
// Initialize the matrix by -1
for(int i = 0; i < 26; i++)
{
for(int j = 0; j < 2; j++)
arr[i][j] = -1;
}
// Initialize the alphabet array
alpha = new ArrayList<>();
for(int i = 0; i < 26; i++)
{
alpha.add(0);
}
answer = new ArrayList<>();
}
// Function to do Depth first
// search Traversal
void dfs(int v)
{
// Pushing current character
answer.add(v);
alpha.set(v, 1);
for(int i = 0; i < 2; i++)
{
if (arr[v][i] == -1 ||
alpha.get(arr[v][i]) == 1)
continue;
dfs(arr[v][i]);
}
}
// Function to find the minimum
// length string
void minString(String str)
{
// Condition if given string
// length is 1
if (str.length() == 1)
{
System.out.println(str);
return;
}
boolean flag = true;
// Loop to find the adjacency
// list of the given string
for(int i = 0; i < str.length() - 1; i++)
{
int j = str.charAt(i) - 'a';
int k = str.charAt(i + 1) - 'a';
// Condition if character
// already present
if (arr[j][0] == k || arr[j][1] == k)
{}
else if (arr[j][0] == -1)
arr[j][0] = k;
else if (arr[j][1] == -1)
arr[j][1] = k;
// Condition if a character
// have more than two different
// adjacent characters
else
{
flag = false;
break;
}
if (arr[k][0] == j || arr[k][1] == j)
{}
else if (arr[k][0] == -1)
arr[k][0] = j;
else if (arr[k][1] == -1)
arr[k][1] = j;
// Condition if a character
// have more than two different
// adjacent characters
else
{
flag = false;
break;
}
}
// Variable to check string contain
// two end characters or not
boolean contain_ends = false;
int count = 0;
int index = 0;
for(int i = 0; i < 26; i++)
{
// Condition if a character has
// only one type of adjacent
// character
if ((arr[i][0] == -1 && arr[i][1] != -1) ||
(arr[i][1] == -1 && arr[i][0] != -1))
{
count++;
index = i;
}
// Condition if the given string
// has exactly two characters
// having only one type of adjacent
// character
if (count == 2)
contain_ends = true;
if (count == 3)
contain_ends = false;
}
if (contain_ends == false || flag == false)
{
System.out.println("Impossible");
return;
}
// Depth first Search Traversal
// on one of the possible end
// character of the string
dfs(index);
// Loop to print the answer
for(int i = 0; i < answer.size(); i++)
{
char ch = (char)(answer.get(i) + 'a');
System.out.print(ch);
}
}
}
class GFG{
// Driver Code
public static void main(String[] args)
{
String s = "acabpba";
Graph graph = new Graph();
graph.minString(s);
}
}
// This code is contributed by sanjeev2552 Python3
# Python implementation to find the
# minimum length string such that
# adjacent characters of the string
# remains adjacent in the string
# Initialize the matrix by -1
arr = [[-1 for i in range(2)] for j in range(26)]
# Initialize the alphabet array
alpha = [0 for i in range(26)]
answer = []
# Function to do Depth first
# search Traversal
def dfs(v):
global arr
global alpha
global answer
# Pushing current character
answer.append(v)
alpha[v] = 1
for i in range(2):
if(alpha[arr[v][i]] == 1 or arr[v][i] == -1):
continue
dfs(arr[v][i])
# Function to find the minimum
# length string
def minString(Str):
# Condition if given string
# length is 1
if(len(Str) == 1):
print(Str)
return
flag = True
temp = 0
# Loop to find the adjacency
# list of the given string
for i in range(0,len(Str) - 1):
j = ord(Str[i]) - ord('a')
k = ord(Str[i + 1]) - ord('a')
# Condition if character
# already present we can do nothing
if(arr[j][0] == k or arr[j][1] == k):
temp = 1
elif(arr[j][0] == -1):
arr[j][0] = k
elif(arr[j][1] == -1):
arr[j][1] = k
# Condition if a character
# have more than two different
# adjacent characters
else:
flag = alse
break
if(arr[k][0] == j or arr[k][1] == j):
temp = 0
elif(arr[k][0] == -1):
arr[k][0] = j
elif(arr[k][1] == -1):
arr[k][1] = j
# Condition if a character
# have more than two different
# adjacent characters
else:
flag = False
break
# Variable to check string contain
# two end characters or not
contain_ends = False
count = 0
index = 0
for i in range(26):
# Condition if a character has
# only one type of adjacent
# character
if((arr[i][0] == -1 and arr[i][1] != -1) or (arr[i][1] == -1 and arr[i][0] != -1)):
count += 1
index = i
# Condition if the given string
# has exactly two characters
# having only one type of adjacent
# character
if(count == 2):
contain_ends = True
if(count == 3):
contain_ends = False
if(contain_ends == False or flag == False):
print("Impossible")
return
# Depth first Search Traversal
# on one of the possible end
# character of the string
dfs(index)
# Loop to print the answer
for i in range(len(answer)):
ch = chr(answer[i] + ord('a'))
print(ch, end = "")
# Driver Code
s = "acabpba"
minString(s)
# This code is contributed by avanitrachhadiya2155输出：
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