找到没有重复字符的给定字符串的后缀数组

给定一个大小为N的字符串str ，任务是找到给定字符串的后缀数组。

注意：后缀数组是给定字符串的所有后缀的排序数组。

例子：

Input: str = “prince”
Output: 4 5 2 3 0 1
Explanation: The suffixes are
0 prince                                    4 ce
1 rince      Sort the suffixes       5 e  
2 ince         —————->    2 ince  
3 nce          alphabetically        3 nce    
4 ce                                          0 prince
5 e                                           1 rince

Input: str = “abcd”
Output: 0 1 2 3

方法：这里讨论任何字符串的后缀数组查找方法。在本文中，重点是为没有重复字符的字符串查找后缀数组。这是一个简单的基于实现的问题。请按照以下步骤解决问题：

• 计算每个字符的出现次数。
• 找到它的前缀和。
• 通过 start[0] = 0, start[i+1] = prefix[i] 查找所有 i > 0 的起始数组。
• 查找包含子字符串（后缀）索引的起始数组，其中包含相应列的起始字符。

请按照下图更好地理解。

插图：

Consider string “prince”:
Given below is the suffix table 

char	a	b	c	d	e	f	g	h	i	j	k	l	m	n	o	p	q	r	s	t	u	v	w	x	y	z
count	0	0	1	0	1	0	0	0	1	0	0	0	0	1	0	1	0	1	0	0	0	0	0	0	0	0
prefix	0	0	1	1	2	2	2	2	3	3	3	3	3	4	4	5	5	6	6	6	6	6	6	6	6	6
start	0	0	0	1	1	2	2	2	2	3	3	3	3	3	4	4	5	5	6	6	6	6	6	6	6	6

For char ‘r’ start value is 5 . 
Implies substring (suffix) starting with char ‘r’ i.e. “rince” has rank 5 .
Rank is position in suffix array.  ( 1 “rince” ) implies 5th position in suffix array ), refer first table.
Similarly, start value of char ‘n’ is 3 . Implies ( 3 “nce” ) 3rd position in suffix array .

下面是上述方法的实现

C++

// C++ code to implement above approach
#include 
using namespace std;
 
// Function to calculate the suffix array
void suffixArray(string str, int N)
{
    // arr[] is array to count
    // occurrence of each character
    int arr[30] = { 0 };
    for (int i = 0; i < N; i++) {
        arr[str[i] - 'a']++;
    }
 
    // Finding prefix count of character
    for (int i = 1; i < 30; i++) {
        arr[i] = arr[i] + arr[i - 1];
    }
 
    int start[30];
    start[0] = 0;
    for (int i = 0; i < 29; i++) {
        start[i + 1] = arr[i];
    }
 
    int ans[N] = { 0 };
 
    // Iterating string in reverse order
    for (int i = N - 1; i >= 0; i--) {
 
        // Storing suffix array in ans[]
        ans[start[str[i] - 'a']] = i;
    }
 
    for (int i = 0; i < N; i++)
        cout << ans[i] << " ";
}
 
// Driver code
int main()
{
    string str = "prince";
    int N = str.length();
 
    suffixArray(str, N);
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
 
 
  // Function to calculate the suffix array
  static void suffixArray(String str, int N)
  {
 
    // arr[] is array to count
    // occurrence of each character
    int arr[] = new int[30];
    for (int i = 0; i < N; i++) {
      arr[str.charAt(i) - 'a']++;
    }
 
    // Finding prefix count of character
    for (int i = 1; i < 30; i++) {
      arr[i] = arr[i] + arr[i - 1];
    }
 
    int start[] = new int[30];
    start[0] = 0;
    for (int i = 0; i < 29; i++) {
      start[i + 1] = arr[i];
    }
 
    int ans[]  = new int[N];
 
    // Iterating string in reverse order
    for (int i = N - 1; i >= 0; i--) {
 
      // Storing suffix array in ans[]
      ans[start[str.charAt(i) - 'a']] = i;
    }
 
    for (int i = 0; i < N; i++)
      System.out.print(ans[i] + " ");
  }
 
  // Driver code
  public static void main (String[] args)
  {
    String str = "prince";
    int N = str.length();
 
    suffixArray(str, N);
  }
}
 
// This code is contributed by hrithikgarg03188

Python3

# Python code for the above approach
 
# Function to calculate the suffix array
def suffixArray(str, N):
   
    # arr[] is array to count
    # occurrence of each character
    arr = [0] * 30
    for i in range(N):
        arr[ord(str[i]) - ord('a')] += 1
 
    # Finding prefix count of character
    for i in range(1, 30):
        arr[i] = arr[i] + arr[i - 1]
 
    start = [0] * 30
    start[0] = 0
    for i in range(29):
        start[i + 1] = arr[i]
 
    ans = [0] * N
 
    # Iterating string in reverse order
    for i in range(N - 1, 0, -1):
 
        # Storing suffix array in ans[]
        ans[start[ord(str[i]) - ord('a')]] = i
 
    for i in range(N):
        print(ans[i], end=" ")
 
# Driver code
str = "prince"
N = len(str)
 
suffixArray(str, N)
 
# This code is contributed by gfgking

C#

// C# code to implement above approach
using System;
class GFG
{
   
    // Function to calculate the suffix array
    static void suffixArray(string str, int N)
    {
        // arr[] is array to count
        // occurrence of each character
        int[] arr = new int[30];
        for (int i = 0; i < N; i++) {
            arr[str[i] - 'a']++;
        }
 
        // Finding prefix count of character
        for (int i = 1; i < 30; i++) {
            arr[i] = arr[i] + arr[i - 1];
        }
 
        int[] start = new int[30];
        start[0] = 0;
        for (int i = 0; i < 29; i++) {
            start[i + 1] = arr[i];
        }
 
        int[] ans = new int[N];
 
        // Iterating string in reverse order
        for (int i = N - 1; i >= 0; i--) {
 
            // Storing suffix array in ans[]
            ans[start[str[i] - 'a']] = i;
        }
 
        for (int i = 0; i < N; i++) {
            Console.Write(ans[i]);
            Console.Write(" ");
        }
    }
 
    // Driver code
    public static int Main()
    {
        string str = "prince";
        int N = str.Length;
 
        suffixArray(str, N);
        return 0;
    }
}
 
// This code is contributed by Taranpreet

Javascript

     // JavaScript code for the above approach
 
 
     // Function to calculate the suffix array
     function suffixArray(str, N) {
         // arr[] is array to count
         // occurrence of each character
         let arr = new Array(30).fill(0);
         for (let i = 0; i < N; i++) {
             arr[str[i].charCodeAt(0) - 'a'.charCodeAt(0)]++;
         }
 
         // Finding prefix count of character
         for (let i = 1; i < 30; i++) {
             arr[i] = arr[i] + arr[i - 1];
         }
 
         let start = new Array(30)
         start[0] = 0;
         for (let i = 0; i < 29; i++) {
             start[i + 1] = arr[i];
         }
 
         let ans = new Array(N).fill(0)
 
         // Iterating string in reverse order
         for (let i = N - 1; i >= 0; i--) {
 
             // Storing suffix array in ans[]
             ans[start[str[i].charCodeAt(0) - 'a'.charCodeAt(0)]] = i;
         }
 
         for (let i = 0; i < N; i++)
             document.write(ans[i] + " ")
     }
 
     // Driver code
 
     let str = "prince";
     let N = str.length;
 
     suffixArray(str, N);
 
 
      // This code is contributed by Potta Lokesh
 

输出

4 5 2 3 0 1 

时间复杂度： O(N)
辅助空间： O(N)