按字典顺序查找位于给定两个字符串之间的字符串

给定两个字符串S 和 T，找到一个长度相同的字符串，按字典顺序大于 S 且小于 T。如果没有形成这样的字符串，则打印“-1”。（S > T）

注意：字符串S = s1s2…sn 据说在字典上小于字符串T = t1t2…tn，如果存在 i，则 s1 = t1, s2 = t2, … si – 1 = ti – 1, si < ti .

例子：

Input : S = "aaa", T = "ccc"
Output : aab
Explanation: 
Here, 'b' is greater than any 
letter in S[]('a') and smaller 
than any letter in T[]('c').

Input : S = "abcde", T = "abcdf"
Output : -1
Explanation: 
There is no other string between
S and T.

方法：找到一个字典序上大于字符串S 的字符串，并检查它是否小于字符串T，如果是则打印下一个字符串，否则打印“-1”。
要查找字符串，以相反的顺序迭代字符串S，如果最后一个字母不是 'z'，则将字母加一（移动到下一个字母）。如果是“z”，则将其更改为“a”并移动到倒数第二个字符。
将结果字符串与字符串T 进行比较，如果两个字符串相等则打印 '-1'，否则打印结果字符串。

以下是上述方法的实现：

C++

// CPP program to find the string
// in lexicographic order which is
// in between given two strings
#include 
using namespace std;
 
// Function to find the lexicographically 
// next string
string lexNext(string s, int n)
{  
    // Iterate from last character
    for (int i = n - 1; i >= 0; i--)
    {  
        // If not 'z', increase by one
        if (s[i] != 'z')
        {
            s[i]++;
            return s;
        }
         
        // if 'z', change it to 'a'
        s[i] = 'a';
    }
}
 
// Driver Code
int main()
{
    string S = "abcdeg", T = "abcfgh";
    int n = S.length();
    string res = lexNext(S, n);
 
    // If not equal, print the
    // resultant string
    if (res != T)
        cout << res << endl;   
    else
        cout << "-1" << endl;
    return 0;
}

Java

//Java program to find the string
// in lexicographic order which is
// in between given two strings
 
class GFG {
 
// Function to find the lexicographically 
// next string
    static String lexNext(String str, int n) {
        char[] s = str.toCharArray();
        // Iterate from last character
        for (int i = n - 1; i >= 0; i--) {
            // If not 'z', increase by one
            if (s[i] != 'z') {
                s[i]++;
                return String.valueOf(s);
            }
 
            // if 'z', change it to 'a'
            s[i] = 'a';
        }
        return null;
    }
 
// Driver Code
    static public void main(String[] args) {
        String S = "abcdeg", T = "abcfgh";
        int n = S.length();
        String res = lexNext(S, n);
 
        // If not equal, print the
        // resultant string
        if (res != T) {
            System.out.println(res);
        } else {
            System.out.println("-1");
        }
    }
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to find the string
# in lexicographic order which is
# in between given two strings
 
# Function to find the lexicographically
# next string
def lexNext(s, n):
 
    # Iterate from last character
    for i in range(n - 1, -1, -1):
 
        # If not 'z', increase by one
        if s[i] != 'z':
            k = ord(s[i])
            s[i] = chr(k + 1)
            return ''.join(s)
 
        # if 'z', change it to 'a'
        s[i] = 'a'
 
# Driver Code
if __name__ == "__main__":
    S = "abcdeg"
    T = "abcfgh"
    n = len(S)
 
    S = list(S)
    res = lexNext(S, n)
 
    # If not equal, print the
    # resultant string
    if res != T:
        print(res)
    else:
        print(-1)
 
# This code is contributed by
# sanjeev2552

C#

//C# program to find the string
// in lexicographic order which is
// in between given two strings
using System;
 
public class GFG {
  
// Function to find the lexicographically 
// next string
    static String lexNext(String str, int n) {
        char[] s = str.ToCharArray();
        // Iterate from last character
        for (int i = n - 1; i >= 0; i--) {
            // If not 'z', increase by one
            if (s[i] != 'z') {
                s[i]++;
                return new String(s);
            }
  
            // if 'z', change it to 'a'
            s[i] = 'a';
        }
        return null;
    }
  
// Driver Code
    static public void Main() {
        String S = "abcdeg", T = "abcfgh";
        int n = S.Length;
        String res = lexNext(S, n);
  
        // If not equal, print the
        // resultant string
        if (res != T) {
            Console.Write(res);
        } else {
            Console.Write("-1");
        }
    }
}
  
// This code is contributed by 29AjayKumar

Javascript

输出：

abcdeh