给定字符串S，任务是更改不遵循下面给出的任何规则的字符串id，并打印更新的字符串。校对的规则是：
- 如果连续三个字符，则表示拼写错误。删除其中一个字符。例如，字符串“ ooops”可以更改为“ oops” 。
- 如果将两对相同的字符(AABB)连接在一起，则表示拼写错误。删除第二对字符之一。例如，字符串“ helloo”可以更改为“ hello” 。
- 规则遵循从左到右的优先级。
例子：
Input: S = “helloo”
Output: hello
Explanation:
As per the Rule #2
helloo => hello
Input: S = “woooow”
Output: woow
Explanation:
As per the Rule #2
woooow => wooow
As per the Rule #1
wooow => woow
方法：想法是遍历字符串，如果拼写错误，请根据给定条件删除多余的字符。由于错误的优先级是从左到右，并且根据给定的规则，可以看出，拼写错误的判断不会发生冲突。考虑从左到右遍历，将已经合法的字符添加到结果中。步骤如下：
- 初始化堆栈以存储字符并比较字符串的最后字符。
- 遍历字符串并将字符添加到堆栈中。
- 检查堆栈的最后3个字符(如果相同)，然后在堆栈顶部弹出该字符。
- 检查堆栈的最后4个字符(如果相同)，然后在堆栈顶部弹出该字符。
- 最后，返回堆栈的字符。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to proofread the spells
string proofreadSpell(string& str)
{
vector result;
// Loop to iterate over the
// characters of the string
for (char c : str) {
// Push the current character c
// in the stack
result.push_back(c);
int n = result.size();
// Check for Rule 1
if (n >= 3) {
if (result[n - 1]
== result[n - 2]
&& result[n - 1]
== result[n - 3]) {
result.pop_back();
}
}
n = result.size();
// Check for Rule 2
if (n >= 4) {
if (result[n - 1]
== result[n - 2]
&& result[n - 3]
== result[n - 4]) {
result.pop_back();
}
}
}
// To store the resultant string
string resultStr = "";
// Loop to iterate over the
// characters of stack
for (char c : result) {
resultStr += c;
}
// Return the resultant string
return resultStr;
}
// Driver Code
int main()
{
// Given string str
string str = "helloo";
// Function Call
cout << proofreadSpell(str);
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to proofread the spells
public static String proofreadSpell(String str)
{
Vector result = new Vector();
// Loop to iterate over the
// characters of the string
for(int i = 0; i < str.length(); i++)
{
// Push the current character c
// in the stack
result.add(str.charAt(i));
int n = result.size();
// Check for Rule 1
if (n >= 3)
{
if (result.get(n - 1) ==
result.get(n - 2) &&
result.get(n - 1) ==
result.get(n - 3))
{
result.remove(result.size() - 1);
}
}
n = result.size();
// Check for Rule 2
if (n >= 4)
{
if (result.get(n - 1) ==
result.get(n - 2) &&
result.get(n - 3) ==
result.get(n - 4))
{
result.remove(result.size() - 1);
}
}
}
// To store the resultant string
String resultStr = "";
// Loop to iterate over the
// characters of stack
for(Character c : result)
{
resultStr += c;
}
// Return the resultant string
return resultStr;
}
// Driver Code
public static void main(String[] args)
{
// Given string str
String str = "helloo";
// Function call
System.out.println(proofreadSpell(str));
}
}
// This code is contributed by divyeshrabadiya07 Python3
# Python3 program for
# the above approach
# Function to proofread
# the spells
def proofreadSpell(str):
result = []
# Loop to iterate over the
# characters of the string
for c in str:
# Push the current character c
# in the stack
result.append(c)
n = len(result)
# Check for Rule 1
if(n >= 3):
if(result[n - 1] == result[n - 2] and
result[n - 1] == result[n - 3]):
result.pop()
n = len(result)
# Check for Rule 2
if(n >= 4):
if(result[n - 1] == result[n - 2] and
result[n - 3] == result[n - 4]):
result.pop()
# To store the
# resultant string
resultStr = ""
# Loop to iterate over the
# characters of stack
for c in result:
resultStr += c
# Return the resultant string
return resultStr
# Driver Code
# Given string str
str = "helloo"
# Function Call
print(proofreadSpell(str))
# This code is contributed by avanitrachhadiya2155C#
// C# program for the above approach
using System;
using System.Collections;
using System.Collections.Generic;
class GFG{
// Function to proofread the spells
static string proofreadSpell(string str)
{
// ArrayList result=new ArrayList();
List result = new List();
// Loop to iterate over the
// characters of the string
foreach(char c in str)
{
// Push the current character c
// in the stack
result.Add(c);
int n = result.Count;
// Check for Rule 1
if (n >= 3)
{
if (result[n - 1] == result[n - 2] &&
result[n - 1] == result[n - 3])
{
result.RemoveAt(n - 1);
}
}
n = result.Count;
// Check for Rule 2
if (n >= 4)
{
if (result[n - 1] == result[n - 2] &&
result[n - 3] == result[n - 4])
{
result.RemoveAt(n - 1);
}
}
}
// To store the resultant string
string resultStr = "";
// Loop to iterate over the
// characters of stack
foreach(char c in result)
{
resultStr += c;
}
// Return the resultant string
return resultStr;
}
// Driver code
public static void Main(string[] args)
{
// Given string str
string str = "helloo";
// Function call
Console.Write(proofreadSpell(str));
}
}
// This code is contributed by rutvik_56 输出：
hello
时间复杂度： O(N)
辅助空间： O(N)