给定两个长度分别为N和M 的排序字符串S1和S2 ，任务是通过合并两个给定的字符串而不改变字符的出现顺序，按字典顺序构造可能的最小字符串。
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”
朴素的方法：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，以获得按字典顺序排列的最小字符串。
时间复杂度： O(N + M)*log(N + M))
辅助空间： O(N + M)
高效的方法：上述方法可以通过使用两点技术通过比较给定字符串的字符来优化，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，比如ptr1和ptr2 ，分别指向字符串S1和S2的开头。
- 初始化一个字符串ans = “”，以存储结果字典中最小的字符串。
- 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 。增加ptr1 。
- 否则，将字符S2[ptr2]附加到字符串ans 。增加ptr2 。
- 完成上述步骤后，其中一个指针没有到达字符串的末尾。
- 因此，将字符串的所有剩余字符添加到字符串ans的末尾。
- 打印ans作为结果字符串。
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]
# Print the 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.Javascript
输出：
aabcdcdhilptx时间复杂度： O(N + M)
辅助空间： O(N + M)
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