给定一个字符串str和一个整数K，任务是通过应用以下操作任意次数来减少字符串，直到不再可能：
Choose a group of K consecutive identical characters and remove them from the string.
最后，打印减少的字符串。
例子：
Input: K = 2, str = “geeksforgeeks”
Output: gksforgks
Explanation: After removal of both occurrences of the substring “ee”, the string reduces to “gksforgks”.
Input: K = 3, str = “qddxxxd”
Output: q
Explanation:
Removal of “xxx” modifies the string to “qddd”.
Again, removal of “ddd”modifies the string to “q”.
方法：这个问题可以使用Stack数据结构来解决。请按照以下步骤解决问题：
- 初始化一个pair
- 遍历字符串的字符。
- 如果当前字符是从字符本目前在堆栈的顶部不同，则设置其频率为1。
- 否则，如果当前字符与堆栈顶部的字符相同，则将其频率增加1 。
- 如果堆栈顶部字符的频率为K ，则将其从堆栈中弹出。
- 最后，将堆栈中剩余的字符打印为结果字符串。
下面是上述方法的实现：
C++
// CPP program for the above approach
#include
#include
using namespace std;
// Basic Approach is to create a Stack that store the
// Character and its continuous repetition number This is
// done using pair Further we check at each
// iteration, whether the character matches the top of stack
// if it does then check for number of repetitions
// else add to top of stack with count 1
class Solution {
public:
string remove_k_char(int k, string s)
{
// Base Case
// If k=1 then all characters
// can be removed at each
// instance
if (k == 1)
return "";
// initialize string
string output = "";
// create a stack using pair<> for storing each
// character and corresponding
// repetition
stack > stk;
// iterate through the string
for (int i = 0; i < s.length(); i++) {
// if stack is empty then simply add the
// character with count 1 else check if
// character is same as top of stack
if (stk.empty() == true) {
stk.push(make_pair(s[i], 1));
}
else {
// if character at top of stack is same as
// current character increase the number of
// repetitions in the top of stack by 1
if (s[i] == (stk.top()).first) {
stk.push(
{ s[i], stk.top().second + 1 });
if (stk.top().second == k) {
int x = k;
while (x) {
stk.pop();
x--;
}
}
}
else {
// if character at top of stack is not
// same as current character push the
// character along with count 1 into the
// top of stack
stk.push(make_pair(s[i], 1));
}
}
}
// Iterate through the stack
// Use string(int,char) in order to replicate the
// character multiple times and convert into string
// then add in front of output string
while (!stk.empty()) {
output += stk.top().first;
stk.pop();
}
reverse(output.begin(), output.end());
return output;
}
};
// Driver Code
int main()
{
string s = "geeksforgeeks";
int k = 2;
Solution obj;
cout << obj.remove_k_char(k, s) << "\n";
return 0;
}
// This code has been contributed by Navansh Goel Java
// Java implementation of the approach
import java.util.*;
class GFG {
// Function to find the reduced string
public static String reduced_String(int k, String s)
{
// Base Case
if (k == 1) {
String ans = "";
return ans;
}
// Creating a stack of type Pair
Stack st = new Stack();
// Length of the string S
int l = s.length();
int ctr = 0;
// iterate through the string
for (int i = 0; i < l; i++) {
// if stack is empty then simply add the
// character with count 1 else check if
// character is same as top of stack
if (st.size() == 0) {
st.push(new Pair(s.charAt(i), 1));
continue;
}
// if character at top of stack is same as
// current character increase the number of
// repetitions in the top of stack by 1
if (st.peek().c == s.charAt(i)) {
Pair p = st.peek();
st.pop();
p.ctr += 1;
if (p.ctr == k) {
continue;
}
else {
st.push(p);
}
}
else {
// if character at top of stack is not
// same as current character push the
// character along with count 1 into the
// top of stack
st.push(new Pair(s.charAt(i), 1));
}
}
// iterate through the stack
// Use string(int,char) in order to replicate the
// character multiple times and convert into string
// then add in front of output string
String ans = "";
while (st.size() > 0) {
char c = st.peek().c;
int cnt = st.peek().ctr;
while (cnt-- > 0)
ans = c + ans;
st.pop();
}
return ans;
}
// Driver code
public static void main(String[] args)
{
int k = 2;
String st = "geeksforgeeks";
String ans = reduced_String(k, st);
System.out.println(ans);
}
static class Pair {
char c;
int ctr;
Pair(char c, int ctr)
{
this.c = c;
this.ctr = ctr;
}
}
} C#
// C# implementation of the above approach
using System;
using System.Collections.Generic;
public class GFG {
// Function to find the reduced string
public static String reduced_String(int k, String s)
{
// Base Case
if (k == 1) {
return "";
}
// Creating a stack of type Pair
Stack st = new Stack();
// Length of the string S
int l = s.Length;
// iterate through the string
for (int i = 0; i < l; i++) {
// if stack is empty then simply add the
// character with count 1 else check if
// character is same as top of stack
if (st.Count == 0) {
st.Push(new Pair(s[i], 1));
continue;
}
// if character at top of stack is same as
// current character increase the number of
// repetitions in the top of stack by 1
if (st.Peek().c == s[i]) {
Pair p = st.Peek();
st.Pop();
p.ctr += 1;
if (p.ctr == k) {
continue;
}
else {
st.Push(p);
}
}
else {
// if character at top of stack is not
// same as current character push the
// character along with count 1 into the
// top of stack
st.Push(new Pair(s[i], 1));
}
}
// iterate through the stack
// Use string(int,char) in order to replicate the
// character multiple times and convert into string
// then add in front of output string
String ans = "";
while (st.Count > 0) {
char c = st.Peek().c;
int cnt = st.Peek().ctr;
while (cnt-- > 0)
ans = c + ans;
st.Pop();
}
return ans;
}
// Driver code
public static void Main(String[] args)
{
int k = 2;
String st = "geeksforgeeks";
String ans = reduced_String(k, st);
Console.Write(ans);
}
public class Pair {
public char c;
public int ctr;
public Pair(char c, int ctr)
{
this.c = c;
this.ctr = ctr;
}
}
}
// This code has been contributed by 29AjayKumar Javascript
C++
#include
using namespace std;
string remove(string s, int k)
{
// count variable to store the count of each character
int cnt = 1, n = s.size();
// to store the output string
string res = "";
// to store the char value
char c = s[0];
for (int i = 1; i <= n; i++) {
if (s[i] == s[i - 1]) {
cnt++;
if (cnt == k) {
i++;
c = s[i];
cnt = 1;
}
}
else {
res += (cnt * c);
c = s[i];
cnt = 1;
}
}
return res;
}
int main()
{
string s = "aabbccd";
int k = 2;
cout << remove(s, k) << endl;
return 0;
} 输出
gksforgks时间复杂度： O(N)
辅助空间： O(N)
方法二：
做法：思路是将count和char值存储在变量中，而不是使用栈来提高空间复杂度
C++
#include
using namespace std;
string remove(string s, int k)
{
// count variable to store the count of each character
int cnt = 1, n = s.size();
// to store the output string
string res = "";
// to store the char value
char c = s[0];
for (int i = 1; i <= n; i++) {
if (s[i] == s[i - 1]) {
cnt++;
if (cnt == k) {
i++;
c = s[i];
cnt = 1;
}
}
else {
res += (cnt * c);
c = s[i];
cnt = 1;
}
}
return res;
}
int main()
{
string s = "aabbccd";
int k = 2;
cout << remove(s, k) << endl;
return 0;
}
输出
d时间复杂度：O(N)
辅助空间： O(1)