查找两个字符串的不常见字符|集合 2
给定两个字符串， str1 和 str2 ，任务是在不使用额外空间的情况下，按排序顺序查找并打印两个给定字符串的不常见字符。在这里，一个不常见的字符意味着该字符要么出现在一个字符串中，要么出现在另一个字符串中，但不能同时出现在两个字符串中。字符串只包含小写字符，可以包含重复字符。
示例:
输入: str1 =“字符”，str2 =“字母” T3】输出: b c l p r
输入: str1 =“极客之怪”，str2 =“极客之怪” T3】输出: f i o q r u z
方法:使用散列的方法已经在这里讨论过了。这个问题也可以通过使用位操作来解决。 该方法使用 2 个变量，存储每个字符的 ASCII 码为-97 的左移位 1 的位或，即 0 代表“a”，1 代表“b”，依此类推。对于这两个字符串，我们在执行这些逐位操作后得到一个整数。现在，这两个整数的异或运算将只在表示不常见字符的位置给出二进制位 1。打印这些位置的字符值。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to print the uncommon
// characters in the given string
// in sorted order
void printUncommon(string str1, string str2)
{
int a1 = 0, a2 = 0;
for (int i = 0; i < str1.length(); i++) {
// Converting character to ASCII code
int ch = int(str1[i]) - 'a';
// Bit operation
a1 = a1 | (1 << ch);
}
for (int i = 0; i < str2.length(); i++) {
// Converting character to ASCII code
int ch = int(str2[i]) - 'a';
// Bit operation
a2 = a2 | (1 << ch);
}
// XOR operation leaves only uncommon
// characters in the ans variable
int ans = a1 ^ a2;
int i = 0;
while (i < 26) {
if (ans % 2 == 1) {
cout << char('a' + i);
}
ans = ans / 2;
i++;
}
}
// Driver code
int main()
{
string str1 = "geeksforgeeks";
string str2 = "geeksquiz";
printUncommon(str1, str2);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG
{
// Function to print the uncommon
// characters in the given string
// in sorted order
static void printUncommon(String str1, String str2)
{
int a1 = 0, a2 = 0;
for (int i = 0; i < str1.length(); i++)
{
// Converting character to ASCII code
int ch = (str1.charAt(i)) - 'a';
// Bit operation
a1 = a1 | (1 << ch);
}
for (int i = 0; i < str2.length(); i++)
{
// Converting character to ASCII code
int ch = (str2.charAt(i)) - 'a';
// Bit operation
a2 = a2 | (1 << ch);
}
// XOR operation leaves only uncommon
// characters in the ans variable
int ans = a1 ^ a2;
int i = 0;
while (i < 26)
{
if (ans % 2 == 1)
{
System.out.print((char) ('a' + i));
}
ans = ans / 2;
i++;
}
}
// Driver code
public static void main(String[] args)
{
String str1 = "geeksforgeeks";
String str2 = "geeksquiz";
printUncommon(str1, str2);
}
}
// This code contributed by Rajput-Ji
C
// C# implementation of the approach
using System;
class GFG
{
// Function to print the uncommon
// characters in the given string
// in sorted order
static void printUncommon(string str1, string str2)
{
int a1 = 0, a2 = 0;
for (int i = 0; i < str1.Length; i++)
{
// Converting character to ASCII code
int ch = (str1[i] - 'a');
// Bit operation
a1 = a1 | (1 << ch);
}
for (int i = 0; i < str2.Length; i++)
{
// Converting character to ASCII code
int ch = (str2[i] - 'a');
// Bit operation
a2 = a2 | (1 << ch);
}
// XOR operation leaves only uncommon
// characters in the ans variable
int ans = a1 ^ a2;
int j = 0;
while (j < 26)
{
if (ans % 2 == 1)
{
Console.Write((char)('a' + j));
}
ans = ans / 2;
j++;
}
}
// Driver code
public static void Main()
{
string str1 = "geeksforgeeks";
string str2 = "geeksquiz";
printUncommon(str1, str2);
}
}
// This code is contributed by SoM15242
Python 3
# Python3 implementation of the approach
# Function to print the uncommon
# characters in the given string
# in sorted order
def printUncommon(str1, str2) :
a1 = 0; a2 = 0;
for i in range(len(str1)) :
# Converting character to ASCII code
ch = ord(str1[i]) - ord('a');
# Bit operation
a1 = a1 | (1 << ch);
for i in range(len(str2)) :
# Converting character to ASCII code
ch = ord(str2[i]) - ord('a');
# Bit operation
a2 = a2 | (1 << ch);
# XOR operation leaves only uncommon
# characters in the ans variable
ans = a1 ^ a2;
i = 0;
while (i < 26) :
if (ans % 2 == 1) :
print(chr(ord('a') + i),end="");
ans = ans // 2;
i += 1;
# Driver code
if __name__ == "__main__" :
str1 = "geeksforgeeks";
str2 = "geeksquiz";
printUncommon(str1, str2);
# This code is contributed by AnkitRai01
java 描述语言
<script>
// Javascript implementation of the approach
// Function to print the uncommon
// characters in the given string
// in sorted order
function printUncommon(str1, str2)
{
var a1 = 0, a2 = 0;
for (var i = 0; i < str1.length; i++) {
// Converting character to ASCII code
var ch = (str1[i].charCodeAt(0)) - 'a'.charCodeAt(0);
// Bit operation
a1 = a1 | (1 << ch);
}
for (var i = 0; i < str2.length; i++) {
// Converting character to ASCII code
var ch = (str2[i].charCodeAt(0)) - 'a'.charCodeAt(0);
// Bit operation
a2 = a2 | (1 << ch);
}
// XOR operation leaves only uncommon
// characters in the ans variable
var ans = a1 ^ a2;
var i = 0;
while (i < 26) {
if (ans % 2 == 1) {
document.write( String.fromCharCode('a'.charCodeAt(0) + i));
}
ans = parseInt(ans / 2);
i++;
}
}
// Driver code
var str1 = "geeksforgeeks";
var str2 = "geeksquiz";
printUncommon(str1, str2);
</script>
Output:
fioqruz
时间复杂度:O(|str1| + |str2| + 26)辅助空间:O(1)
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