包含两个字符串中不常见字符的连接字符串
原文：https://www.geeksforgeeks.org/concatenated-string-uncommon-characters-two-strings/
给出了两个字符串，您必须修改第一个字符串，以便必须删除第二个字符串的所有常见字符，并且第二个字符串的不常见字符必须与第一个字符串的不常见字符连接在一起。
示例：
Input : S1 = "aacdb"
S2 = "gafd"
Output : "cbgf"
Input : S1 = "abcs";
S2 = "cxzca";
Output : "bsxz"
这个想法是使用哈希映射，其中键是字符，值是存在字符的字符串数。 如果一个字符存在于一个字符串中，则计数为 1，否则，如果字符存在于两个字符串中，则计数为 2。
-
将结果初始化为空字符串。
-
将计数为 1 的第二个字符串的所有字符推送到映射中。
-
遍历第一个字符串，并将所有这些字符附加到映射中不存在的结果中。 映射中显示的字符数为 2。
-
遍历第二个字符串并附加所有这些字符，得出计数为 1 的结果。
C++
// C++ program Find concatenated string with
// uncommon characters of given strings
#include <bits/stdc++.h>
using namespace std;
string concatenetedString(string s1, string s2)
{
string res = ""; // result
// store all characters of s2 in map
unordered_map<char, int> m;
for (int i = 0; i < s2.size(); i++)
m[s2[i]] = 1;
// Find characters of s1 that are not
// present in s2 and append to result
for (int i = 0; i < s1.size(); i++) {
if (m.find(s1[i]) == m.end())
res += s1[i];
else
m[s1[i]] = 2;
}
// Find characters of s2 that are not
// present in s1\.
for (int i = 0; i < s2.size(); i++)
if (m[s2[i]] == 1)
res += s2[i];
return res;
}
/* Driver program to test above function */
int main()
{
string s1 = "abcs";
string s2 = "cxzca";
cout << concatenetedString(s1, s2);
return 0;
}
Java
// Java program Find concatenated string with
// uncommon characters of given strings
import java.util.*;
import java.lang.*;
import java.io.*;
class gfg {
public static String concatenatedString(String s1, String s2)
{
// Result
String res = "";
int i;
// creating a hashMap to add characters in string s2
HashMap<Character, Integer> m = new HashMap<Character, Integer>();
for (i = 0; i < s2.length(); i++)
m.put(s2.charAt(i), 1);
// Find characters of s1 that are not
// present in s2 and append to result
for (i = 0; i < s1.length(); i++)
if (!m.containsKey(s1.charAt(i)))
res += s1.charAt(i);
else
m.put(s1.charAt(i), 2);
// Find characters of s2 that are not
// present in s1\.
for (i = 0; i < s2.length(); i++)
if (m.get(s2.charAt(i)) == 1)
res += s2.charAt(i);
return res;
}
// Driver code
public static void main(String[] args)
{
String s1 = "abcs";
String s2 = "cxzca";
System.out.println(concatenatedString(s1, s2));
}
}
/* This code is contributed by Devarshi_Singh*/
Python 3
# Python3 program Find concatenated string
# with uncommon characters of given strings
def concatenetedString(s1, s2):
res = "" # result
m = {}
# store all characters of s2 in map
for i in range(0, len(s2)):
m[s2[i]] = 1
# Find characters of s1 that are not
# present in s2 and append to result
for i in range(0, len(s1)):
if(not s1[i] in m):
res = res + s1[i]
else:
m[s1[i]] = 2
# Find characters of s2 that are not
# present in s1.
for i in range(0, len(s2)):
if(m[s2[i]] == 1):
res = res + s2[i]
return res
# Driver Code
if __name__ == "__main__":
s1 = "abcs"
s2 = "cxzca"
print(concatenetedString(s1, s2))
# This code is contributed
# by Sairahul099
C
// C# program Find concatenated string with
// uncommon characters of given strings
using System;
using System.Collections.Generic;
class GFG
{
public static String concatenatedString(String s1,
String s2)
{
// Result
String res = "";
int i;
// creating a hashMap to add characters
// in string s2
Dictionary<char,
int> m = new Dictionary<char,
int>();
for (i = 0; i < s2.Length; i++)
if (!m.ContainsKey(s2[i]))
m.Add(s2[i], 1);
// Find characters of s1 that are not
// present in s2 and append to result
for (i = 0; i < s1.Length; i++)
if (!m.ContainsKey(s1[i]))
res += s1[i];
else
m[s1[i]] = 2;
// Find characters of s2 that are not
// present in s1\.
for (i = 0; i < s2.Length; i++)
if (m[s2[i]] == 1)
res += s2[i];
return res;
}
// Driver code
public static void Main(String[] args)
{
String s1 = "abcs";
String s2 = "cxzca";
Console.WriteLine(concatenatedString(s1, s2));
}
}
// This code is contributed by PrinciRaj1992
输出：
bsxz
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