给定的排序字符串S选自N小写字符，任务是给定的字符串，使得没有两个相邻的字符是相同的重新排列字符。如果无法按照给定的条件重新排列，则打印“-1” 。
例子：
Input: S = “aaabc”
Output: abaca
Input: S = “aa”
Output: -1
方法：给定的问题可以通过使用两点技术来解决。请按照以下步骤解决此问题：
- 遍历字符串S和检查的字符，如果没有两个相邻的字符字符串中的同一然后打印字符串S。
- 否则，如果字符串的大小为2并且具有相同的字符，则打印“-1” 。
- 初始化三个变量，例如， i为0 ， j为1 ， k为2以遍历字符串S 。
- 当k小于N 时迭代并执行以下步骤：
- 如果S [i]不等于S [j]时，然后通过递增1 i和j，以及增量K内由1，如果j的值等于k。
- 否则，如果S [j]的等于S [k]的，由1个增量ķ。
- 否则，交换s[j]和s[k]并将i和j增加1 ，如果j等于k，则将k增加1 。
- 完成上述步骤后，反转字符串S 。
- 最后，遍历字符串S的字符并检查是否没有两个相邻的字符相同。如果发现为真，则打印字符串S 。否则，打印“-1” 。
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 string S
// contains pair of adjacent
// characters that are equal or not
bool isAdjChar(string& s)
{
// Traverse the string S
for (int i = 0; i < s.size() - 1; i++) {
// If S[i] and S[i+1] are equal
if (s[i] == s[i + 1])
return true;
}
// Otherwise, return false
return false;
}
// Function to rearrange characters
// of a string such that no pair of
// adjacent characters are the same
void rearrangeStringUtil(string& S, int N)
{
// Initialize 3 variables
int i = 0, j = 1, k = 2;
// Iterate until k < N
while (k < N) {
// If S[i] is not equal
// to S[j]
if (S[i] != S[j]) {
// Increment i and j by 1
i++;
j++;
// If j equals k and increment
// the value of K by 1
if (j == k) {
k++;
}
}
// Else
else {
// If S[j] equals S[k]
if (S[j] == S[k]) {
// Increment k by 1
k++;
}
// Else
else {
// Swap
swap(S[k], S[j]);
// Increment i and j
// by 1
i++;
j++;
// If j equals k
if (j == k) {
// Increment k by 1
k++;
}
}
}
}
}
// Function to rearrange characters
// in a string so that no two
// adjacent characters are same
string rearrangeString(string& S, int N)
{
// If string is already valid
if (isAdjChar(S) == false) {
return S;
}
// If size of the string is 2
if (S.size() == 2)
return "-1";
// Function Call
rearrangeStringUtil(S, N);
// Reversing the string
reverse(S.begin(), S.end());
// Function Call
rearrangeStringUtil(S, N);
// If the string is valid
if (isAdjChar(S) == false) {
return S;
}
// Otherwise
return "-1";
}
// Driver Code
int main()
{
string S = "aaabc";
int N = S.length();
cout << rearrangeString(S, N);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG
{
static char []S = "aaabc".toCharArray();
// Function to check if a String S
// contains pair of adjacent
// characters that are equal or not
static boolean isAdjChar()
{
// Traverse the String S
for (int i = 0; i < S.length - 1; i++)
{
// If S[i] and S[i+1] are equal
if (S[i] == S[i + 1])
return true;
}
// Otherwise, return false
return false;
}
// Function to rearrange characters
// of a String such that no pair of
// adjacent characters are the same
static void rearrangeStringUtil(int N)
{
// Initialize 3 variables
int i = 0, j = 1, k = 2;
// Iterate until k < N
while (k < N) {
// If S[i] is not equal
// to S[j]
if (S[i] != S[j]) {
// Increment i and j by 1
i++;
j++;
// If j equals k and increment
// the value of K by 1
if (j == k) {
k++;
}
}
// Else
else {
// If S[j] equals S[k]
if (S[j] == S[k]) {
// Increment k by 1
k++;
}
// Else
else {
// Swap
swap(k,j);
// Increment i and j
// by 1
i++;
j++;
// If j equals k
if (j == k) {
// Increment k by 1
k++;
}
}
}
}
}
static void swap(int i, int j)
{
char temp = S[i];
S[i] = S[j];
S[j] = temp;
}
// Function to rearrange characters
// in a String so that no two
// adjacent characters are same
static String rearrangeString(int N)
{
// If String is already valid
if (isAdjChar() == false) {
return String.valueOf(S);
}
// If size of the String is 2
if (S.length == 2)
return "-1";
// Function Call
rearrangeStringUtil(N);
// Reversing the String
reverse();
// Function Call
rearrangeStringUtil(N);
// If the String is valid
if (isAdjChar() == false) {
return String.valueOf(S);
}
// Otherwise
return "-1";
}
static void reverse() {
int l, r = S.length - 1;
for (l = 0; l < r; l++, r--) {
char temp = S[l];
S[l] = S[r];
S[r] = temp;
}
}
// Driver Code
public static void main(String[] args)
{
int N = S.length;
System.out.print(rearrangeString(N));
}
}
// This code is contributed by Princi SinghPython3
# Python 3 program for the above approach
S = "aaabc"
# Function to check if a string S
# contains pair of adjacent
# characters that are equal or not
def isAdjChar(s):
# Traverse the string S
for i in range(len(s)-1):
# If S[i] and S[i+1] are equal
if (s[i] == s[i + 1]):
return True
# Otherwise, return false
return False
# Function to rearrange characters
# of a string such that no pair of
# adjacent characters are the same
def rearrangeStringUtil(N):
global S
S = list(S)
# Initialize 3 variables
i = 0
j = 1
k = 2
# Iterate until k < N
while (k < N):
# If S[i] is not equal
# to S[j]
if (S[i] != S[j]):
# Increment i and j by 1
i += 1
j += 1
# If j equals k and increment
# the value of K by 1
if (j == k):
k += 1
# Else
else:
# If S[j] equals S[k]
if (S[j] == S[k]):
# Increment k by 1
k += 1
# Else
else:
# Swap
temp = S[k]
S[k] = S[j]
S[j] = temp
# Increment i and j
# by 1
i += 1
j += 1
# If j equals k
if (j == k):
# Increment k by 1
k += 1
S = ''.join(S)
# Function to rearrange characters
# in a string so that no two
# adjacent characters are same
def rearrangeString(N):
global S
# If string is already valid
if (isAdjChar(S) == False):
return S
# If size of the string is 2
if (len(S) == 2):
return "-1"
# Function Call
rearrangeStringUtil(N)
# Reversing the string
S = S[::-1]
# Function Call
rearrangeStringUtil(N)
# If the string is valid
if (isAdjChar(S) == False):
return S
# Otherwise
return "-1"
# Driver Code
if __name__ == '__main__':
N = len(S)
print(rearrangeString(N))
# This code is contributed by ipg2016107.C#
// C# program for the above approach
using System;
public class GFG
{
static char []S = "aaabc".ToCharArray();
// Function to check if a String S
// contains pair of adjacent
// characters that are equal or not
static bool isAdjChar()
{
// Traverse the String S
for (int i = 0; i < S.Length - 1; i++)
{
// If S[i] and S[i+1] are equal
if (S[i] == S[i + 1])
return true;
}
// Otherwise, return false
return false;
}
// Function to rearrange characters
// of a String such that no pair of
// adjacent characters are the same
static void rearrangeStringUtil(int N)
{
// Initialize 3 variables
int i = 0, j = 1, k = 2;
// Iterate until k < N
while (k < N) {
// If S[i] is not equal
// to S[j]
if (S[i] != S[j]) {
// Increment i and j by 1
i++;
j++;
// If j equals k and increment
// the value of K by 1
if (j == k) {
k++;
}
}
// Else
else {
// If S[j] equals S[k]
if (S[j] == S[k]) {
// Increment k by 1
k++;
}
// Else
else {
// Swap
swap(k,j);
// Increment i and j
// by 1
i++;
j++;
// If j equals k
if (j == k) {
// Increment k by 1
k++;
}
}
}
}
}
static void swap(int i, int j)
{
char temp = S[i];
S[i] = S[j];
S[j] = temp;
}
// Function to rearrange characters
// in a String so that no two
// adjacent characters are same
static String rearrangeString(int N)
{
// If String is already valid
if (isAdjChar() == false) {
return String.Join("",S);
}
// If size of the String is 2
if (S.Length == 2)
return "-1";
// Function Call
rearrangeStringUtil(N);
// Reversing the String
reverse();
// Function Call
rearrangeStringUtil(N);
// If the String is valid
if (isAdjChar() == false) {
return String.Join("",S);
}
// Otherwise
return "-1";
}
static void reverse() {
int l, r = S.Length - 1;
for (l = 0; l < r; l++, r--) {
char temp = S[l];
S[l] = S[r];
S[r] = temp;
}
}
// Driver Code
public static void Main(String[] args)
{
int N = S.Length;
Console.Write(rearrangeString(N));
}
}
// This code is contributed by shikhasingrajput输出：
acaba时间复杂度： O(N)
辅助空间： O(1)
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