通过增加每个字符与单词末尾的距离来修改字符串
原文:https://www . geesforgeks . org/modify-string-by-递增-每个字符与其单词末尾的距离/
给定一个字符串 S ，任务是通过用一个新字符替换每个字符 S[i] 来修改给定的字符串，该新字符的值为( S[i] +它在单词末尾的位置)
示例:
输入:S =“ACM fkz” 输出:“CDM hlz” 解释: 给定字符串中有 2 个单词{“ACM”、“fkz”} 对于“ACM”: a 变成‘a’+2 =‘c’ c 变成‘c’+1 =‘d’ m 变成‘m’+0 =‘m’ “ACM” 同样，“fkz”变成“hlz”。 因此，需要的答案是“cdm hlz”
输入:“极客换极客” 输出:“khgls HPR khgls”
方法:思路是把给定的字符串拆分成单词并单独修改每个单词。以下是步骤:
- 首先，将给定的字符串 S 标记为单个单词。
- 迭代单词并为单词中的每个字符添加其从末尾到它的位置。
- 然后，将结果单词添加到最后一个字符串中，说 res 。
- 继续重复以上两个步骤，直到字符串中的每个单词都被转换。
以下是上述方法的程序:
C++
// C++ implementation of
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to transform and return
// the transformed word
string util(string sub)
{
int n = sub.length();
int i = 0;
// Stores resulting word
string ret = "";
// Iterate over the word
while (i < n)
{
// Add the position
// value to the letter
int t = (sub[i] - 'a') +
n - 1 - i;
// Convert it back to character
char ch = (char)(t % 26 + 97);
// Add it to the string
ret = ret + ch;
i++;
}
return ret;
}
// Function to transform the
// given string
void manipulate(string s)
{
// Size of string
int n = s.length();
int i = 0, j = 0;
// Stores resultant string
string res = "";
// Iterate over given string
while (i < n)
{
// End of word is reached
if (s[i] == ' ')
{
// Append the word
res += util(s.substr(j, i));
res = res + " ";
j = i + 1;
i = j + 1;
}
else
{
i++;
}
}
// For the last word
res = res + util(s.substr(j, i));
cout << res << endl;
}
// Driver code
int main()
{
// Given string
string s = "acm fkz";
// Function call
manipulate(s);
return 0;
}
// This code is contributed by divyeshrabadiya07
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of
// the above approach
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG {
// Function to transform the given string
public static void manipulate(String s)
{
// Size of string
int n = s.length();
int i = 0, j = 0;
// Stores resultant string
String res = "";
// Iterate over given string
while (i < n) {
// End of word is reached
if (s.charAt(i) == ' ') {
// Append the word
res += util(s.substring(j, i));
res = res + " ";
j = i + 1;
i = j + 1;
}
else {
i++;
}
}
// For the last word
res = res + util(s.substring(j, i));
System.out.println(res);
}
// Function to transform and return
// the transformed word
public static String util(String sub)
{
int n = sub.length();
int i = 0;
// Stores resulting word
String ret = "";
// Iterate over the word
while (i < n) {
// Add the position
// value to the letter
int t = (sub.charAt(i) - 'a')
+ n - 1 - i;
// Convert it back to character
char ch = (char)(t % 26 + 97);
// Add it to the string
ret = ret + String.valueOf(ch);
i++;
}
return ret;
}
// Driver Code
public static void main(String[] args)
{
// Given string
String s = "acm fkz";
// Function Call
manipulate(s);
}
}
Python 3
# Python3 implementation of
# the above approach
# Function to transform and return
# the transformed word
def util(sub):
n = len(sub)
i = 0
# Stores resulting word
ret = ""
# Iterate over the word
while i < n:
# Add the position
# value to the letter
t = (ord(sub[i]) - 97) + n - 1 - i
# Convert it back to character
ch = chr(t % 26 + 97)
# Add it to the string
ret = ret + ch
i = i + 1
return ret
# Function to transform the
# given string
def manipulate(s):
# Size of string
n = len(s)
i = 0
j = 0
# Stores resultant string
res = ""
# Iterate over given string
while i < n:
# End of word is reached
if s[i] == ' ':
# print(s[j:j+i])
# Append the word
res += util(s[j : j + i])
res = res + " "
j = i + 1
i = j + 1
else:
i = i + 1
# For the last word
res = res + util(s[j : j + i])
print(res)
# Driver code
if __name__ == "__main__":
# Given string
s = "acm fkz"
# Function call
manipulate(s)
# This code is contributed by akhilsaini
C
// C# implementation of
// the above approach
using System;
class GFG{
// Function to transform the given string
public static void manipulate(String s)
{
// Size of string
int n = s.Length;
int i = 0, j = 0;
// Stores resultant string
String res = "";
// Iterate over given string
while (i < n)
{
// End of word is reached
if (s[i] == ' ')
{
// Append the word
res += util(s.Substring(j, i - j));
res = res + " ";
j = i + 1;
i = j + 1;
}
else
{
i++;
}
}
// For the last word
res = res + util(s.Substring(j, i - j));
Console.WriteLine(res);
}
// Function to transform and return
// the transformed word
public static String util(String sub)
{
int n = sub.Length;
int i = 0;
// Stores resulting word
String ret = "";
// Iterate over the word
while (i < n)
{
// Add the position
// value to the letter
int t = (sub[i] - 'a') +
n - 1 - i;
// Convert it back to character
char ch = (char)(t % 26 + 97);
// Add it to the string
ret = ret + String.Join("", ch);
i++;
}
return ret;
}
// Driver Code
public static void Main(String[] args)
{
// Given string
String s = "acm fkz";
// Function Call
manipulate(s);
}
}
// This code is contributed by shikhasingrajput
Output:
cdm hlz
时间复杂度:O(N) T5辅助空间:** O(N)
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