给定由小写字母组成的字符串str ,任务是找到为字符串 的每个字符获得的最大重复字符。如果对于任何索引,超过一个字符出现的次数达到最大值,则打印最近出现的字符。
例子:
Input: str = “abbc”
Output:
a -> 1
b -> 1
b -> 2
b -> 2
Explanation:
str[0] = ‘a’. Therefore, print a -> 1.
str[1] = ‘b’. Now ‘a’ and ‘b’ have equal frequency. Since, ‘b’ is the most recently occurring character, print b -> 1.
str[2] = ‘b’. Since ‘b’is the most repeating character, print b -> 2.
str[3] = ‘c’. Since ‘b’is the most repeating character, print b -> 2.
Input: str = “htdddg”
Output:h -> 1
t -> 1
d -> 1
d -> 2
d -> 3
d -> 3
方法:按照下面给出的步骤解决问题:
- 初始化一个数组,比如freq[]来存储字符串中每个不同字符的频率。
- 初始化两个变量,比如max , charMax分别存储最大重复字符和最大重复字符的频率。
- 遍历字符串并更新当前字符的频率,更新max和charMax的值。
- 最后,在每个字符遍历后打印charMax和max的值。
下面是上述方法的实现:
C++
// C++ Program to implement
// the above approach
#include
using namespace std;
// Function to print the
// maximum repeating
// character at each index
// of the String
void findFreq(string str,
int N)
{
// Stores frequency of
// each distinct character
int freq[256];
memset(freq, 0,
sizeof(freq));
// Stores frequency of
// maximum repeating
// character
int max = 0;
// Stores the character having
// maximum frequency
char charMax = '0';
// Traverse the String
for (int i = 0; i < N; i++)
{
// Stores current character
char ch = str[i];
// Update the frequency of str[i]
freq[ch]++;
// If frequency of current
// character exceeds max
if (freq[ch] >= max)
{
// Update max
max = freq[ch];
// Update charMax
charMax = ch;
}
// Print the required output
cout<< charMax << "->" <<
max << endl;
}
}
// Driver Code
int main()
{
string str = "abbc";
// Stores length of str
int N = str.size();
findFreq(str, N);
}
// This code is contributed by Rajput-Ji
Java
// Java program to implement
// the above approach
import java.util.*;
public class GFG {
// Function to print the maximum repeating
// character at each index of the string
public static void findFreq(String str,
int N)
{
// Stores frequency of
// each distinct character
int[] freq = new int[256];
// Stores frequency of maximum
// repeating character
int max = 0;
// Stores the character having
// maximum frequency
char charMax = '0';
// Traverse the string
for (int i = 0; i < N; i++) {
// Stores current character
char ch = str.charAt(i);
// Update the frequency of str[i]
freq[ch]++;
// If frequency of current
// character exceeds max
if (freq[ch] >= max) {
// Update max
max = freq[ch];
// Update charMax
charMax = ch;
}
// Print the required output
System.out.println(
charMax + " -> " + max);
}
}
// Driver Code
public static void main(String[] args)
{
String str = "abbc";
// Stores length of str
int N = str.length();
findFreq(str, N);
}
}
Python3
# Python3 program to implement
# the above approach
# Function to print the maximum repeating
# character at each index of the strring
def findFreq(strr, N):
# Stores frequency of
# each distinct character
freq = [0] * 256
# Stores frequency of maximum
# repeating character
max = 0
# Stores the character having
# maximum frequency
charMax = '0'
# Traverse the strring
for i in range(N):
# Stores current character
ch = ord(strr[i])
# Update the frequency of strr[i]
freq[ch] += 1
# If frequency of current
# character exceeds max
if (freq[ch] >= max):
# Update max
max = freq[ch]
# Update charMax
charMax = ch
# Print the required output
print(chr(charMax), "->", max)
# Driver Code
if __name__ == '__main__':
strr = "abbc"
# Stores length of strr
N = len(strr)
findFreq(strr, N)
# This code is contributed by mohit kumar 29
C#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to print the maximum repeating
// character at each index of the string
public static void findFreq(String str,
int N)
{
// Stores frequency of
// each distinct character
int[] freq = new int[256];
// Stores frequency of maximum
// repeating character
int max = 0;
// Stores the character having
// maximum frequency
char charMax = '0';
// Traverse the string
for (int i = 0; i < N; i++)
{
// Stores current character
char ch = str[i];
// Update the frequency of
// str[i]
freq[ch]++;
// If frequency of current
// character exceeds max
if (freq[ch] >= max)
{
// Update max
max = freq[ch];
// Update charMax
charMax = ch;
}
// Print the required output
Console.WriteLine(charMax +
" -> " + max);
}
}
// Driver Code
public static void Main(String[] args)
{
String str = "abbc";
// Stores length of str
int N = str.Length;
findFreq(str, N);
}
}
// This code is contributed by shikhasingrajput
输出:
a -> 1
b -> 1
b -> 2
b -> 2
时间复杂度:O(N)
辅助空间: O(1)
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