给定一个字符串str由字符和数字和整数k的，任务是解密字符串和在经解密的字符串返回第k个字符。
为了解密字符串，通过字符遍历字符串字符，如果当前字符是字母，然后将其追加到别的结果字符串，如果它是那么数字位解析的数量和重复得到的字符串的时候，这解析的数量和继续原始字符串。例如，str =“ ab2c3”将被解密为“ ababcababcababc”。
例子：
Input: str = “ab2c3”, k = 5
Output: c
Decrypted string will be “ababcababcababc” and ‘c’ is the fifth character.
Input: str = “x2y3”, k = 3
Output: y
方法：
- 初始化起始索引i = 0和total_len = 0 。
- 当i小于输入字符串的长度时循环播放，并检查当前字符是否为字母。如果是，则将total_len递增1，并检查总长度是否小于或等于k(如果是)，然后返回字符串，否则递增i 。
- 初始化n = 0再次循环，而i小于输入字符串的长度且i不是字母，然后解析数字并递增i并找到next_total_len = total_len * n。
- 如果k
- 如果找不到位置，则更新position = total_len ，最后返回第k个字符。如果找不到，则返回-1 。
- 如果k
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
#include
using namespace std;
// Function to print kth character of
// String s after decrypting it
char findKthChar(string s, int k)
{
// Get the length of string
int len = s.length();
// Initialise pointer to character
// of input string to zero
int i = 0;
// Total length of resultant string
int total_len = 0;
// Traverse the string from starting
// and check if each character is
// alphabet then increment total_len
while (i < len) {
if (isalpha(s[i])) {
total_len++;
// If total_leg equal to k then
// return string else increment i
if (total_len == k)
return s[i];
i++;
}
else {
// Parse the number
int n = 0;
while (i < len && !isalpha(s[i])) {
n = n * 10 + (s[i] - '0');
i++;
}
// Update next_total_len
int next_total_len = total_len * n;
// Get the position of kth character
if (k <= next_total_len) {
int pos = k % total_len;
// Position not found then update
// position with total_len
if (!pos) {
pos = total_len;
}
// Recursively find the kth position
return findKthChar(s, pos);
}
else {
// Else update total_len
// by next_total_len
total_len = next_total_len;
}
}
}
// Return -1 if character not found
return -1;
}
// Driver code
int main()
{
string s = "ab2c3";
int k = 5;
cout << findKthChar(s, k);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GfG
{
// Function to print kth character of
// String s after decrypting it
static Character findKthChar(String s, int k)
{
// Get the length of string
int len = s.length();
// Initialise pointer to character
// of input string to zero
int i = 0;
// Total length of resultant string
int total_len = 0;
// Traverse the string from starting
// and check if each character is
// alphabet then increament total_len
while (i < len)
{
if (Character.isLetter(s.charAt(i)))
{
total_len++;
// If total_leg equal to k then
// return string else increment i
if (total_len == k)
return s.charAt(i);
i++;
}
else
{
// Parse the number
int n = 0;
while (i < len && !Character.isLetter(s.charAt(i)))
{
n = n * 10 + (s.charAt(i) - '0');
i++;
}
// Update next_total_len
int next_total_len = total_len * n;
// Get the position of kth character
if (k <= next_total_len)
{
int pos = k % total_len;
// Position not found then update
// position with total_len
if (pos == 0)
{
pos = total_len;
}
// Recursively find the kth position
return findKthChar(s, pos);
}
else
{
// Else update total_len
// by next_total_len
total_len = next_total_len;
}
}
}
// Return -1 if character not found
return ' ';
}
// Driver code
public static void main(String[] args)
{
String s = "ab2c3";
int k = 5;
System.out.println(findKthChar(s, k));
}
}
// This code is contributed by Prerna Saini.Python3
# Python 3 implementation of the approach
# Function to print kth character of
# String s after decrypting it
def findKthChar(s, k):
# Get the length of string
len1 = len(s)
# Initialise pointer to character
# of input string to zero
i = 0
# Total length of resultant string
total_len = 0
# Traverse the string from starting
# and check if each character is
# alphabet then increament total_len
while (i < len1):
if (s[i].isalpha()):
total_len += 1
# If total_leg equal to k then
# return string else increment i
if (total_len == k):
return s[i]
i += 1
else:
# Parse the number
n = 0
while (i < len1 and s[i].isalpha() == False):
n = n * 10 + (ord(s[i]) - ord('0'))
i += 1
# Update next_total_len
next_total_len = total_len * n
# Get the position of kth character
if (k <= next_total_len):
pos = k % total_len
# Position not found then update
# position with total_len
if (pos == 0):
pos = total_len
# Recursively find the kth position
return findKthChar(s, pos)
else:
# Else update total_len
# by next_total_len
total_len = next_total_len
# Return -1 if character not found
return -1
# Driver code
if __name__ == '__main__':
s = "ab2c3"
k = 5
print(findKthChar(s, k))
# This code is contributed by
# Surendra_GangwarC#
// C# implementation of the approach
using System;
class GFG
{
// Function to print kth character of
// String s after decrypting it
static char findKthChar(String s, int k)
{
// Get the length of string
int len = s.Length;
// Initialise pointer to character
// of input string to zero
int i = 0;
// Total length of resultant string
int total_len = 0;
// Traverse the string from starting
// and check if each character is
// alphabet then increament total_len
while (i < len)
{
if (char.IsLetter(s[i]))
{
total_len++;
// If total_leg equal to k then
// return string else increment i
if (total_len == k)
return s[i];
i++;
}
else
{
// Parse the number
int n = 0;
while (i < len && !char.IsLetter(s[i]))
{
n = n * 10 + (s[i] - '0');
i++;
}
// Update next_total_len
int next_total_len = total_len * n;
// Get the position of kth character
if (k <= next_total_len)
{
int pos = k % total_len;
// Position not found then update
// position with total_len
if (pos == 0)
{
pos = total_len;
}
// Recursively find the kth position
return findKthChar(s, pos);
}
else
{
// Else update total_len
// by next_total_len
total_len = next_total_len;
}
}
}
// Return -1 if character not found
return ' ';
}
// Driver code
public static void Main(String[] args)
{
String s = "ab2c3";
int k = 5;
Console.WriteLine(findKthChar(s, k));
}
}
// This code is contributed by PrinciRaj1992输出：
c
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