通过一次添加一个字符来合并两个字符串，从而在字典上尽可能大

给定两个字符串S和T，任务是通过从任一字符串的开头一次添加一个字符来合并这两个字符串，以形成结果字符串。结果字符串应该是字典序最大的字符串 可以通过合并字符串S和T 形成。

例子：

Input: S = “dbcbb”, T = “cdbbb”
Output : dcdbcbbbbb

Input : S = “geeks“, T = “forgeeks”
Output : gforgeekseeks

方法：解决问题的最简单的想法是从字典中比其他字符串大的字符串贪婪地选择第一个字符。因此，可以使用贪心算法和递归来解决这个问题。请按照以下步骤解决问题：

•  如果任一字符串长度为0 ，则返回S + T作为答案。
• 如果S在字典上大于T，则返回S[0] + maximumMerge(S.substr(1), T) 。
• 否则，取T的第一个字符并调用递归函数maximumMerge(S, T.substr(1))。

下面是上述方法的实现：

C++

// C++ program for the approach
 
#include 
using namespace std;
 
// Recursive  bfunction for finding
// the lexicographically largest string
string largestMerge(string s1, string s2)
{
    // If either of the string length is 0,
    // return the other string
    if (s1.size() == 0 || s2.size() == 0)
        return s1 + s2;
 
    // If s1 is lexicographically
    // larger than s2
    if (s1 > s2) {
 
        // Take first character of s1
        // and call the function
        return s1[0] + largestMerge(s1.substr(1), s2);
    }
 
    // Take first character of s2
    // and recursively call function for
    // remaining string
    return s2[0] + largestMerge(s1, s2.substr(1));
}
 
// Driver Code
int main()
{
    // Given Input
    string s1 = "geeks";
    string s2 = "forgeeks";
 
    // Function Call
    cout << largestMerge(s1, s2) << endl;
 
    return 0;
}

Python3

# Python program for the above approach
 
# Recursive function for finding
# the lexicographically largest string
def largestMerge(s1, s2):
 
    # If either of the string length is 0,
    # return the other string
    if len(s1) == 0 or len(s2) == 0:
        return s1+s2
 
    # If s1 is lexicographically
    # larger than s2
    if(s1 > s2):
       
        # Take first character of s1
        # and call the function
        return s1[0]+largestMerge(s1[1:], s2)
 
    # Take first character of s2
    # and recursively call function for
    # remaining string
    return s2[0]+largestMerge(s1, s2[1:])
 
 
# Driver code
if __name__ == '__main__':
   
    # Given Input
    s1 = "geeks"
    s2 = "forgeeks"
 
    # Function call
    print(largestMerge(s1, s2))
     
# This code is contributed by MuskanKalra1

C#

// C# program for the approach
using System;
class GFG {
    // Recursive  bfunction for finding
    // the lexicographically largest string
    static string largestMerge(string s1, string s2)
    {
        // If either of the string length is 0,
        // return the other string
        if (s1.Length == 0 || s2.Length == 0)
            return s1 + s2;
 
        // If s1 is lexicographically
        // larger than s2
        if (string.Compare(s1, s2) == 1) {
 
            // Take first character of s1
            // and call the function
            return s1[0]
                + largestMerge(s1.Substring(1), s2);
        }
 
        // Take first character of s2
        // and recursively call function for
        // remaining string
        return s2[0] + largestMerge(s1, s2.Substring(1));
    }
 
    // Driver Code
    public static void Main()
    {
        // Given Input
        string s1 = "geeks";
        string s2 = "forgeeks";
 
        // Function Call
        Console.Write(largestMerge(s1, s2));
    }
}
 
// This code is contributed by ukasp.

输出

gforgeekseeks

时间复杂度： O(M×N)，其中M和N分别是字符串s1和s2的长度。
辅助空间： O(1)

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