通过仅删除单一类型的字符来检查给定的字符串是否可以成为回文
给定一个字符串S ，任务是是否可以制作一个字符串 删除相同字符的出现后的回文，任意次数
例子：
Input: S = “abczdzacb”
Output: Yes
Explanation: Remove first and second occurrence of character ‘a’, string S becomes “bczdzcb”, which is a palindrome .
Input: S = “madem”
Output: No
方法：该任务可以通过迭代给定字符串中的每个唯一字符来解决，并在不匹配的地方删除它的出现，如果找到有效的回文，在多次删除相同字符的出现后，返回“是” 否则返回“否”。
请按照以下步骤解决问题：
- 开始迭代字符串的每个唯一字符，其出现将被删除
- 使用双指针技术检查不匹配，将l放在字符串的开头，将r放在字符串的结尾
- 如果S[l] == S[r] ，增加l ，减少r 。
- 如果S[l]!= S[r]，检查是否 S[l[ == char，执行 l++，否则如果 S[r] == char，执行 r–
- 如果条件都不成立，则表示给定不能转换为回文
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to check if a palindrome is
// possible or not
string isPossible(string S)
{
// Stores the length of string
int n = (int)S.length();
// Stores the unique characters in
// the string
set st;
for (int i = 0; i < n; i++) {
st.insert(S[i]);
}
// Check if valid palindrome is
// possible or not
bool check = false;
// Iterating over unique characters
// of the string
for (auto ele : st) {
// Pointers to check the condition
int low = 0, high = n - 1;
bool flag = true;
// Iterating over the string
for (int i = 0; i < n; i++) {
if (S[low] == S[high]) {
// Updating low and high
low++;
high--;
}
else {
if (S[low] == ele) {
// Updating low
low++;
}
else if (S[high] == ele) {
// Updating high
high--;
}
else {
// It is impossible
// to make palindrome
// by removing
// occurrences of char
flag = false;
break;
}
}
}
// If palindrome is formed
// break the loop
if (flag == true) {
check = true;
break;
}
}
if (check)
return "Yes";
else
return "No";
}
// Driver Code
int main()
{
string S = "abczdzacb";
cout << isPossible(S);
return 0;
} Java
// Java code for the above approach
import java.util.*;
class GFG
{
// Function to check if a palindrome is
// possible or not
static String isPossible(String S)
{
// Stores the length of string
int n = S.length();
// Stores the unique characters in
// the string
Set st = new HashSet();
for (int i = 0; i < n; i++) {
st.add(S.charAt(i));
}
// Check if valid palindrome is
// possible or not
boolean check = false;
// Iterating over unique characters
// of the string
for (Character ele : st) {
// Pointers to check the condition
int low = 0, high = n - 1;
boolean flag = true;
// Iterating over the string
for (int i = 0; i < n; i++) {
if (S.charAt(low) == S.charAt(high)) {
// Updating low and high
low++;
high--;
}
else {
if (S.charAt(low) == ele) {
// Updating low
low++;
}
else if (S.charAt(high)== ele) {
// Updating high
high--;
}
else {
// It is impossible
// to make palindrome
// by removing
// occurrences of char
flag = false;
break;
}
}
}
// If palindrome is formed
// break the loop
if (flag == true) {
check = true;
break;
}
}
if (check)
return "Yes";
else
return "No";
}
// Driver Code
public static void main (String[] args) {
String S = "abczdzacb";
System.out.println(isPossible(S));
}
}
// This code is contributed by Potta Lokesh Python3
# python program for the above approach
# Function to check if a palindrome is
# possible or not
def isPossible(S):
# Stores the length of string
n = len(S)
# Stores the unique characters in
# the string
st = set()
for i in range(0, n):
st.add(S[i])
# Check if valid palindrome is
# possible or not
check = False
# Iterating over unique characters
# of the string
for ele in st:
# Pointers to check the condition
low = 0
high = n - 1
flag = True
# Iterating over the string
for i in range(0, n):
if (S[low] == S[high]):
# Updating low and high
low += 1
high -= 1
else:
if (S[low] == ele):
# Updating low
low += 1
elif (S[high] == ele):
# Updating high
high -= 1
else:
# It is impossible
# to make palindrome
# by removing
# occurrences of char
flag = False
break
# If palindrome is formed
# break the loop
if (flag == True):
check = True
break
if (check):
return "Yes"
else:
return "No"
# Driver Code
if __name__ == "__main__":
S = "abczdzacb"
print(isPossible(S))
# This code is contributed by rakeshsahniC#
// C# code for the above approach
using System;
using System.Collections.Generic;
public class GFG
{
// Function to check if a palindrome is
// possible or not
static String isPossible(String S)
{
// Stores the length of string
int n = S.Length;
// Stores the unique characters in
// the string
HashSet st = new HashSet();
for (int i = 0; i < n; i++) {
st.Add(S[i]);
}
// Check if valid palindrome is
// possible or not
bool check = false;
// Iterating over unique characters
// of the string
foreach (char ele in st) {
// Pointers to check the condition
int low = 0, high = n - 1;
bool flag = true;
// Iterating over the string
for (int i = 0; i < n; i++) {
if (S[low] == S[high]) {
// Updating low and high
low++;
high--;
}
else {
if (S[low] == ele) {
// Updating low
low++;
}
else if (S[high]== ele) {
// Updating high
high--;
}
else {
// It is impossible
// to make palindrome
// by removing
// occurrences of char
flag = false;
break;
}
}
}
// If palindrome is formed
// break the loop
if (flag == true) {
check = true;
break;
}
}
if (check)
return "Yes";
else
return "No";
}
// Driver Code
public static void Main(String[] args) {
String S = "abczdzacb";
Console.WriteLine(isPossible(S));
}
}
// This code is contributed by shikhasingrajput Javascript
输出：
Yes时间复杂度： O(n*26)
辅助空间： 在）