给定两个分别具有长度N和M的分类字符串S1和S2 ，任务是通过合并两个给定的字符串而不改变字符出现的顺序来构造词典上最小的字符串。
例子：
Input: S1 = “eefgkors”, S2 = “eegks”
Output: “eeeefggkkorss”
Explanation:
String “eeeefggkkorss” is lexicographically the smallest string that can be formed after merging the two given string S1 and S2.
Input: S1 = “aabcdtx”, S2 = “achilp”
Output: “aaabccdhilptx”
天真的方法：最简单的方法是将给定的两个字符串连接起来，并对结果字符串进行排序，以按字典顺序获得最小的字符串。
时间复杂度： O(N + M)* log(N + M))
辅助空间： O(N + M)
高效的方法：可以通过使用两指针技术通过比较给定字符串的字符来优化上述方法，然后相应地构造所需的字符串。
请按照以下步骤解决问题：
- 初始化两个指针ptr1和ptr2 ，分别指向字符串S1和S2的开头。
- 初始化字符串ans =“”，以存储结果按字典顺序最小的字符串。
- 迭代直到ptr1小于N且ptr2小于M并执行以下步骤：
- 如果S1 [ptr1]小于S2 [ptr2] ，则将字符S1 [ptr1]附加到字符串ans上。递增ptr1 。
- 否则，将字符S2 [ptr2]附加到字符串ans上。递增ptr2 。
- 完成上述步骤后，指针之一未到达字符串的末尾。
- 因此，将字符串的所有剩余字符添加到字符串ans的末尾。
- 将ans打印为结果字符串。
下面是上述方法的实现：
C++14
// C++ program for the above approach
#include
using namespace std;
// Function to find lexicographically
// smallest string possible by merging
// two sorted strings
void mergeStrings(string s1, string s2)
{
// Stores length of string s1
int len1 = s1.size();
// Stores length of string s2
int len2 = s2.size();
// Pointer to beginning
// of string1 i.e., s1
int pntr1 = 0;
// Pointer to beginning
// of string2 i.e., s2
int pntr2 = 0;
// Stores the final string
string ans = "";
// Traverse the strings
while (pntr1 < len1 && pntr2 < len2) {
// Append the smaller of the
// two current characters
if (s1[pntr1] < s2[pntr2]) {
ans += s1[pntr1];
pntr1++;
}
else {
ans += s2[pntr2];
pntr2++;
}
}
// Append the remaining characters
// of any of the two strings
if (pntr1 < len1) {
ans += s1.substr(pntr1, len1);
}
if (pntr2 < len2) {
ans += s2.substr(pntr2, len2);
}
// Print the final string
cout << ans;
}
// Driver Code
int main()
{
string S1 = "abdcdtx";
string S2 = "achilp";
// Function Call
mergeStrings(S1, S2);
return 0;
} Java
// Java program for the above approach
import java.io.*;
class GFG{
// Function to find lexicographically
// smallest string possible by merging
// two sorted strings
static void mergeStrings(String s1, String s2)
{
// Stores length of string s1
int len1 = s1.length();
// Stores length of string s2
int len2 = s2.length();
// Pointer to beginning
// of string1 i.e., s1
int pntr1 = 0;
// Pointer to beginning
// of string2 i.e., s2
int pntr2 = 0;
// Stores the final string
String ans = "";
// Traverse the strings
while (pntr1 < len1 && pntr2 < len2)
{
// Append the smaller of the
// two current characters
if (s1.charAt(pntr1) < s2.charAt(pntr2))
{
ans += s1.charAt(pntr1);
pntr1++;
}
else
{
ans += s2.charAt(pntr2);
pntr2++;
}
}
// Append the remaining characters
// of any of the two strings
if (pntr1 < len1)
{
ans += s1.substring(pntr1, len1);
}
if (pntr2 < len2)
{
ans += s2.substring(pntr2, len2);
}
// Print the final string
System.out.println(ans);
}
// Driver Code
public static void main (String[] args)
{
String S1 = "abdcdtx";
String S2 = "achilp";
// Function Call
mergeStrings(S1, S2);
}
}
// This code is contributed by sanjoy_62Python3
# Python3 program for the above approach
# Function to find lexicographically
# smallest possible by merging
# two sorted strings
def mergeStrings(s1, s2):
# Stores length of s1
len1 = len(s1)
# Stores length of s2
len2 = len(s2)
# Pointer to beginning
# of string1 i.e., s1
pntr1 = 0
# Pointer to beginning
# of string2 i.e., s2
pntr2 = 0
# Stores the final string
ans = ""
# Traverse the strings
while (pntr1 < len1 and pntr2 < len2):
# Append the smaller of the
# two current characters
if (s1[pntr1] < s2[pntr2]):
ans += s1[pntr1]
pntr1 += 1
else:
ans += s2[pntr2]
pntr2 += 1
# Append the remaining characters
# of any of the two strings
if (pntr1 < len1):
ans += s1[pntr1:pntr1 + len1]
if (pntr2 < len2):
ans += s2[pntr2:pntr2 + len2]
# Prthe final string
print (ans)
# Driver Code
if __name__ == '__main__':
S1 = "abdcdtx"
S2 = "achilp"
# Function Call
mergeStrings(S1, S2)
# This code is contributed by mohit kumar 29.C#
// C# program for the above approach
using System;
class GFG
{
// Function to find lexicographically
// smallest string possible by merging
// two sorted strings
static void mergeStrings(string s1, string s2)
{
// Stores length of string s1
int len1 = s1.Length;
// Stores length of string s2
int len2 = s2.Length;
// Pointer to beginning
// of string1 i.e., s1
int pntr1 = 0;
// Pointer to beginning
// of string2 i.e., s2
int pntr2 = 0;
// Stores the final string
string ans = "";
// Traverse the strings
while (pntr1 < len1 && pntr2 < len2) {
// Append the smaller of the
// two current characters
if (s1[pntr1] < s2[pntr2]) {
ans += s1[pntr1];
pntr1++;
}
else {
ans += s2[pntr2];
pntr2++;
}
}
// Append the remaining characters
// of any of the two strings
if (pntr1 < len1) {
ans += s1.Substring(pntr1, len1 - pntr1);
}
if (pntr2 < len2) {
ans += s2.Substring(pntr2, len2 - pntr2);
}
// Print the final string
Console.WriteLine(ans);
}
// Driver Code
public static void Main()
{
string S1 = "abdcdtx";
string S2 = "achilp";
// Function Call
mergeStrings(S1, S2);
}
}
// This code is contributed by ukasp.输出：
aabcdcdhilptx时间复杂度： O(N + M)
辅助空间： O(N + M)