通过将每个字符向右循环移动各自的频率来修改字符串
给定一个由小写英文字母组成的字符串 S ，任务是将给定字符串 S 的每个字符按其频率循环右移。
字符的循环移动是指将字符“z”移动到“a”，作为其下一个字符。
示例:
输入: S = "geeksforgeeks" 输出: iiimugpsiiimu 说明: 对字符串 S 进行以下更改:
- g 的频率是 2。因此，将字符“g”移动 2 就变成了“I”。
- “e”的频率是 4。因此，将字符“e”移动 4 就变成了“I”。
- “k”的频率是 2。因此，将字符“k”移动 2 就变成了“m”。
- 的频率是 2。因此，将字符的移动 2 就变成了 u。
- “f”的频率为 1。因此，将字符“f”移动 1 就变成了“g”。
- “o”的频率为 1。因此，将字符“o”移动 1 就变成了“p”。
- “r”的频率为 1。因此，将字符“r”移动 1 就变成了“s”。
在上述字符移动之后，字符串修改为“iiimugpsiiimu”。
输入:S = " aabcadb " T3】输出:dddd
方法:解决这个问题的思路是遍历字符串找到字符串中每个字符的出现频率，然后按其频率递增每个字符。按照以下步骤解决问题:
- 初始化一个数组，比如频率[] ，存储字符串 S 中每个字符的出现次数。
- 遍历给定的字符串 S ，并执行以下步骤:
- 求当前字符 S[i] 的频率。
- 增加当前字符的频率，并将 S[i] 的值更新为其更新的字符。
- 完成上述步骤后，打印字符串 S 作为结果修改字符串。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to replace the characters
// by its frequency of character in it
void addFrequencyToCharacter(string S)
{
// Stores frequencies of characters
// in the string S
int frequency[26] = { 0 };
int N = S.length();
// Traverse the string S
for (int i = 0; i < N; i++) {
// Increment the frequency of
// each character by 1
frequency[S[i] - 'a'] += 1;
}
// Traverse the string S
for (int i = 0; i < N; i++) {
// Find the frequency of
// the current character
int add = frequency[S[i] - 'a'] % 26;
// Update the character
if (int(S[i]) + add
<= int('z'))
S[i] = char(int(S[i])
+ add);
else {
add = (int(S[i]) + add)
- (int('z'));
S[i] = char(int('a')
+ add - 1);
}
}
// Print the resultant string
cout << S;
}
// Driver Code
int main()
{
string S = "geeksforgeeks";
addFrequencyToCharacter(S);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
public class GFG {
// Function to replace the characters
// by its frequency of character in it
static void addFrequencyToCharacter(String Str)
{
// Stores frequencies of characters
// in the string S
int frequency[] = new int[26];
int N = Str.length();
char S[] = Str.toCharArray();
// Traverse the string S
for (int i = 0; i < N; i++) {
// Increment the frequency of
// each character by 1
frequency[S[i] - 'a'] += 1;
}
// Traverse the string S
for (int i = 0; i < N; i++) {
// Find the frequency of
// the current character
int add = frequency[S[i] - 'a'] % 26;
// Update the character
if ((int)(S[i]) + add <= (int)('z'))
S[i] = (char)((int)(S[i]) + add);
else {
add = ((int)(S[i]) + add) - ((int)('z'));
S[i] = (char)((int)('a') + add - 1);
}
}
// Print the resultant string
System.out.println(new String(S));
}
// Driver Code
public static void main(String[] args)
{
String S = "geeksforgeeks";
addFrequencyToCharacter(S);
}
}
// This code is contributed by Kingash.
Python 3
# Python3 program for the above approach
# Function to replace the characters
# by its frequency of character in it
def addFrequencyToCharacter(S):
# Stores frequencies of characters
# in the string S
frequency = [0 for i in range(26)]
N = len(S)
S = list(S)
# Traverse the string S
for i in range(N):
# Increment the frequency of
# each character by 1
frequency[ord(S[i]) - ord('a')] += 1
# Traverse the string S
for i in range(N):
# Find the frequency of
# the current character
add = frequency[ord(S[i]) - ord('a')] % 26
# Update the character
if ord(S[i]) + add <= ord('z'):
S[i] = chr(ord(S[i]) + add)
else:
add = ord(S[i]) + add - ord('z')
S[i] = chr(ord('a') + add - 1)
# Print the resultant string
s = ""
print(s.join(S))
# Driver Code
if __name__ == '__main__':
S = "geeksforgeeks"
addFrequencyToCharacter(S)
# This code is contributed by jana_sayantan
C
// C# program for the above approach
using System;
class GFG{
// Function to replace the characters
// by its frequency of character in it
static void addFrequencyToCharacter(string Str)
{
// Stores frequencies of characters
// in the string S
int[] frequency = new int[26];
int N = Str.Length;
char[] S = Str.ToCharArray();
// Traverse the string S
for(int i = 0; i < N; i++)
{
// Increment the frequency of
// each character by 1
frequency[S[i] - 'a'] += 1;
}
// Traverse the string S
for(int i = 0; i < N; i++)
{
// Find the frequency of
// the current character
int add = frequency[S[i] - 'a'] % 26;
// Update the character
if ((int)(S[i]) + add <= (int)('z'))
S[i] = (char)((int)(S[i]) + add);
else
{
add = ((int)(S[i]) + add) - ((int)('z'));
S[i] = (char)((int)('a') + add - 1);
}
}
// Print the resultant string
Console.Write(new string(S));
}
// Driver Code
public static void Main(string[] args)
{
string S = "geeksforgeeks";
addFrequencyToCharacter(S);
}
}
// This code is contributed by ukasp
java 描述语言
<script>
// Javascript program for the above approach
// Function to replace the characters
// by its frequency of character in it
function addFrequencyToCharacter(Str)
{
// Stores frequencies of characters
// in the string S
var frequency = Array.from({length: 26}, (_, i) => 0);
var N = Str.length;
var S = Str.split('');
// Traverse the string S
for (var i = 0; i < N; i++) {
// Increment the frequency of
// each character by 1
frequency[S[i].charCodeAt(0) -
'a'.charCodeAt(0)] += 1;
}
// Traverse the string S
for (var i = 0; i < N; i++) {
// Find the frequency of
// the current character
var add = frequency[S[i].charCodeAt(0) -
'a'.charCodeAt(0)] % 26;
// Update the character
if ((S[i].charCodeAt(0)) +
add <= ('z').charCodeAt(0))
S[i] = String.fromCharCode((S[i].charCodeAt(0))
+ add);
else {
add = ((S[i].charCodeAt(0)) + add) -
(('z').charCodeAt(0));
S[i] = String.fromCharCode(('a'.charCodeAt(0)) +
add - 1);
}
}
// Print the resultant string
document.write(S.join(''));
}
// Driver Code
var S = "geeksforgeeks";
addFrequencyToCharacter(S);
// This code contributed by shikhasingrajput
</script>
Output:
iiimugpsiiimu
时间复杂度:O(N) T5辅助空间:** O(26)
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