给定字符串str ，任务是在按字典顺序排列的所有子字符串中找到给定字符串的等级。
例子：
Input: S = “enren”
Output: 7
Explanation:
All the possible substrings in the sorted order are {“e”, “e”, “en”, “en”, “enr”, “enre”, “enren”, “n”, “n”, “nr”, “nre”, “nren”, “r”, “re”, “ren”}.
Therefore, the rank of the given string “enren” is 7.
Input: S = “geeks”
Output: 12
Explanation:
All possible substrings in the sorted order are {“e”, “e”, “ee”, “eek”, “eeks”, “ek”, “eks”, “g”, “ge”, “gee”, “geek”, “geeks”, “k”, “ks”, “s”}.
Therefore, the rank of the given string “geeks” is 12.
方法：请按照以下步骤解决问题：
- 初始化长度为26的向量的数组arr [] ，以存储字符串存在的字符的索引，并将其排名为0。
- 存储每个字符的索引。的索引将被存储在ARR [0]，对于b，编曲[1]将存储其所有指数等。
- 遍历存储在数组ARR []到其比字符串S的第一字符较小字符的每个索引。
- 对于每个索引i ，从该索引开始的子字符串总数为N – i 。加N – i进行排名，因为具有这些索引的所有字符都较小。
- 现在，遍历之后，存储所有从S的第一个字符开始的子字符串，并按字典顺序对这些子字符串进行排序。
- 遍历排序的子字符串，并将每个子字符串与字符串S进行比较，并递增等级，直到找到等于S的子字符串。
- 打印等级值+1，以获取给定字符串的等级。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find lexiographic rank
// of string among all its substring
int lexiographicRank(string s)
{
// Length of string
int n = s.length();
vector alphaIndex[26];
// Traverse the given string
// and store the indices of
// each character
for (int i = 0; i < s.length(); i++) {
// Extract the index
int x = s[i] - 'a';
// Store it in the vector
alphaIndex[x].push_back(i);
}
// Traverse the aphaIndex array
// lesser than the index of first
// character of given string
int rank = 0;
for (int i = 0; i < 26
&& 'a' + i < s[0];
i++) {
// If alphaIndex[i] size exceeds 0
if (alphaIndex[i].size() > 0) {
// Traverse over the indices
for (int j = 0;
j < alphaIndex[i].size(); j++) {
// Add count of substring
// equal to n - alphaIndex[i][j]
rank = rank
+ (n
- alphaIndex[i][j]);
}
}
}
vector str;
for (int i = 0;
i < alphaIndex[s[0] - 'a'].size();
i++) {
// Store all substrings in a vector
// str starting with the first
// character of the given string
string substring;
int j = alphaIndex[s[0] - 'a'][i];
for (; j < n; j++) {
// Insert the current
// character to substring
substring.push_back(s[j]);
// Store the substring formed
str.push_back(substring);
}
}
// Sort the substring in the
// lexicographical order
sort(str.begin(), str.end());
// Find the rank of given string
for (int i = 0; i < str.size(); i++) {
if (str[i] != s) {
// increase the rank until
// the given string is same
rank++;
}
// If substring is same as
// the given string
else {
break;
}
}
// Add 1 to rank of
// the given string
return rank + 1;
}
// Driver Code
int main()
{
// Given string
string str = "enren";
// Function Call
cout << lexiographicRank(str);
return 0;
} Java
// Java program for
// the above approach
import java.util.*;
class GFG{
// Function to find lexiographic rank
// of String among all its subString
static int lexiographicRank(char []s)
{
// Length of String
int n = s.length;
Vector []alphaIndex = new Vector[26];
for (int i = 0; i < alphaIndex.length; i++)
alphaIndex[i] = new Vector();
// Traverse the given String
// and store the indices of
// each character
for (int i = 0; i < s.length; i++)
{
// Extract the index
int x = s[i] - 'a';
// Store it in the vector
alphaIndex[x].add(i);
}
// Traverse the aphaIndex array
// lesser than the index of first
// character of given String
int rank = 0;
for (int i = 0; i < 26 &&
'a' + i < s[0]; i++)
{
// If alphaIndex[i] size exceeds 0
if (alphaIndex[i].size() > 0)
{
// Traverse over the indices
for (int j = 0;
j < alphaIndex[i].size();
j++)
{
// Add count of subString
// equal to n - alphaIndex[i][j]
rank = rank + (n - alphaIndex[i].get(j));
}
}
}
Vector str = new Vector();
for (int i = 0;
i < alphaIndex[s[0] - 'a'].size();
i++)
{
// Store all subStrings in a vector
// str starting with the first
// character of the given String
String subString = "";
int j = alphaIndex[s[0] - 'a'].get(i);
for (; j < n; j++)
{
// Insert the current
// character to subString
subString += (s[j]);
// Store the subString formed
str.add(subString);
}
}
// Sort the subString in the
// lexicographical order
Collections.sort(str);
// Find the rank of given String
for (int i = 0; i < str.size(); i++)
{
if (!str.get(i).equals(String.valueOf(s)))
{
// increase the rank until
// the given String is same
rank++;
}
// If subString is same as
// the given String
else
{
break;
}
}
// Add 1 to rank of
// the given String
return rank + 1;
}
// Driver Code
public static void main(String[] args)
{
// Given String
String str = "enren";
// Function Call
System.out.print(lexiographicRank(str.toCharArray()));
}
}
// This code is contributed by shikhasingrajput Python3
# Python3 program for the above approach
# Function to find lexiographic rank
# of among all its substrring
def lexiographicRank(s):
# Length of strring
n = len(s)
alphaIndex = [[] for i in range(26)]
# Traverse the given strring
# and store the indices of
# each character
for i in range(len(s)):
# Extract the index
x = ord(s[i]) - ord('a')
# Store it in the vector
alphaIndex[x].append(i)
# Traverse the aphaIndex array
# lesser than the index of first
# character of given strring
rank = -1
for i in range(26):
if ord('a') + i >= ord(s[0]):
break
# If alphaIndex[i] size exceeds 0
if len(alphaIndex[i]) > 0:
# T raverse over the indices
for j in range(len(alphaIndex[i])):
# Add count of substrring
# equal to n - alphaIndex[i][j]
rank = rank + (n - alphaIndex[i][j])
# print(rank)
strr = []
for i in range(len(alphaIndex[ord(s[0]) - ord('a')])):
# Store all substrrings in a vector
# strr starting with the first
# character of the given strring
substrring = []
jj = alphaIndex[ord(s[0]) - ord('a')][i]
for j in range(jj, n):
# Insert the current
# character to substrring
substrring.append(s[j])
# Store the subformed
strr.append(substrring)
# Sort the subin the
# lexicographical order
strr = sorted(strr)
# Find the rank of given strring
for i in range(len(strr)):
if (strr[i] != s):
# Increase the rank until
# the given is same
rank += 1
# If subis same as
# the given strring
else:
break
# Add 1 to rank of
# the given strring
return rank + 1
# Driver Code
if __name__ == '__main__':
# Given strring
strr = "enren"
# Function call
print(lexiographicRank(strr))
# This code is contributed by mohit kumar 29C#
// C# program for
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find lexiographic rank
// of String among all its subString
static int lexiographicRank(char []s)
{
// Length of String
int n = s.Length;
List []alphaIndex = new List[26];
for (int i = 0; i < alphaIndex.Length; i++)
alphaIndex[i] = new List();
// Traverse the given String
// and store the indices of
// each character
for (int i = 0; i < s.Length; i++)
{
// Extract the index
int x = s[i] - 'a';
// Store it in the vector
alphaIndex[x].Add(i);
}
// Traverse the aphaIndex array
// lesser than the index of first
// character of given String
int rank = 0;
for (int i = 0; i < 26 &&
'a' + i < s[0]; i++)
{
// If alphaIndex[i] size exceeds 0
if (alphaIndex[i].Count > 0)
{
// Traverse over the indices
for (int j = 0;
j < alphaIndex[i].Count; j++)
{
// Add count of subString
// equal to n - alphaIndex[i,j]
rank = rank + (n - alphaIndex[i][j]);
}
}
}
List str = new List();
for (int i = 0;
i < alphaIndex[s[0] - 'a'].Count; i++)
{
// Store all subStrings in a vector
// str starting with the first
// character of the given String
String subString = "";
int j = alphaIndex[s[0] - 'a'][i];
for (; j < n; j++)
{
// Insert the current
// character to subString
subString += (s[j]);
// Store the subString formed
str.Add(subString);
}
}
// Sort the subString in the
// lexicographical order
str.Sort();
// Find the rank of given String
for (int i = 0; i < str.Count; i++)
{
if (!str[i].Equals(String.Join("", s)))
{
// increase the rank until
// the given String is same
rank++;
}
// If subString is same as
// the given String
else
{
break;
}
}
// Add 1 to rank of
// the given String
return rank + 1;
}
// Driver Code
public static void Main(String[] args)
{
// Given String
String str = "enren";
// Function Call
Console.Write(lexiographicRank(str.ToCharArray()));
}
}
// This code is contributed by 29AjayKumar 输出：
7
时间复杂度： O(N 3 )
辅助空间： O(N)