给定一个字符串s ，任务是找出使字符串s回文所需的最小相邻交换次数。如果不可能，则返回-1 。
例子：
Input: aabcb
Output: 3
Explanation:
After 1st swap: abacb
After 2nd swap: abcab
After 3rd swap: abcba
Input: adbcdbad
Output: -1
方法
以下是解决此问题的详细步骤。
- 取双指针，其中第一个指针从字符串的左侧开始跟踪，第二个指针从字符串的右侧开始跟踪。
- 直到我们找到相同的字符，继续将右指针向左移动一步。
- 如果未找到相同的字符，则返回 -1。
- 如果找到相同的字符，则将右指针的字符向右交换，直到它不在字符串的正确位置。
- 增加左指针并重复步骤 2。
下面是上述方法的实现：
C++
// C++ program to Count
// minimum swap to make
// string palindrome
#include
using namespace std;
// Function to Count minimum swap
int countSwap(string s)
{
// calculate length of string as n
int n = s.length();
// counter to count minimum swap
int count = 0;
// A loop which run till mid of
// string
for (int i = 0; i < n / 2; i++) {
// Left pointer
int left = i;
// Right pointer
int right = n - left - 1;
// A loop which run from right
// pointer towards left pointer
while (left < right) {
// if both char same then
// break the loop.
// If not, then we have to
// move right pointer to one
// position left
if (s[left] == s[right]) {
break;
}
else {
right--;
}
}
// If both pointers are at same
// position, it denotes that we
// don't have sufficient characters
// to make palindrome string
if (left == right) {
return -1;
}
// else swap and increase the count
for (int j = right; j < n - left - 1;
j++) {
swap(s[j], s[j + 1]);
count++;
}
}
return count;
}
// Driver code
int main()
{
string s = "geeksfgeeks";
// Function calling
int ans1 = countSwap(s);
reverse(s.begin(),s.end());
int ans2 = countSwap(s);
cout< Java
// Java program to Count
// minimum swap to make
// string palindrome
import java.util.*;
class GFG {
// Function to Count minimum swap
static int countSwap(String str)
{
// Legth of string
int n = str.length();
// it will convert string to
// char array
char s[] = str.toCharArray();
// Counter to count minimum
// swap
int count = 0;
// A loop which run in half
// string from starting
for (int i = 0; i < n / 2;
i++) {
// Left pointer
int left = i;
// Right pointer
int right = n - left - 1;
// A loop which run from
// right pointer to left
// pointer
while (left < right) {
// if both char same
// then break the loop
// if not same then we
// have to move right
// pointer to one step
// left
if (s[left] == s[right]) {
break;
}
else {
right--;
}
}
// it denotes both pointer at
// same position and we don't
// have sufficient char to make
// palindrome string
if (left == right) {
return -1;
}
else {
for (int j = right;
j < n - left - 1; j++) {
char t = s[j];
s[j] = s[j + 1];
s[j + 1] = t;
count++;
}
}
}
return count;
}
// Driver Code
public static void main(String[] args)
{
String s = "geeksfgeeks";
// Function calling
int ans1 = countSwap(s);
StringBuilder sb=new StringBuilder(s);
sb.reverse();
s = sb.toString();
int ans2 = countSwap(s);
if(ans1 > ans2)
System.out.println(ans1);
else
System.out.println(ans2);
}
}Python3
''' Python3 program to Count
minimum swap to make
string palindrome'''
# Function to Count minimum swap
def CountSwap(s, n):
s = list(s)
# Counter to count minimum swap
count = 0
ans = True
# A loop which run in half string
# from starting
for i in range(n // 2):
# Left pointer
left = i
# Right pointer
right = n - left - 1
# A loop which run from right pointer
# to left pointer
while left < right:
# if both char same then
# break the loop if not
# same then we have to move
# right pointer to one step left
if s[left] == s[right]:
break
else:
right -= 1
# it denotes both pointer at
# same position and we don't
# have sufficient char to make
# palindrome string
if left == right:
ans = False
break
else:
for j in range(right, n - left - 1):
(s[j], s[j + 1]) = (s[j + 1], s[j])
count += 1
if ans:
return (count)
else:
return -1
# Driver Code
s = 'geeksfgeeks'
# Legth of string
n = len(s)
# Function calling
ans1 = CountSwap(s, n)
ans2 = CountSwap(s[::-1], n)
print(max(ans1, ans2))C#
// C# program to Count
// minimum swap to make
// string palindrome
using System;
class GFG {
// Function to Count minimum swap
static int countSwap(String str)
{
// Length of string
int n = str.Length;
// it will convert string to
// char array
char []s = str.ToCharArray();
// Counter to count minimum
// swap
int count = 0;
// A loop which run in half
// string from starting
for (int i = 0; i < n / 2;
i++) {
// Left pointer
int left = i;
// Right pointer
int right = n - left - 1;
// A loop which run from
// right pointer to left
// pointer
while (left < right) {
// if both char same
// then break the loop
// if not same then we
// have to move right
// pointer to one step
// left
if (s[left] == s[right]) {
break;
}
else {
right--;
}
}
// it denotes both pointer at
// same position and we don't
// have sufficient char to make
// palindrome string
if (left == right) {
return -1;
}
else {
for (int j = right;
j < n - left - 1; j++) {
char t = s[j];
s[j] = s[j + 1];
s[j + 1] = t;
count++;
}
}
}
return count;
}
// Driver Code
public static void Main(String[] args)
{
String s = "geeksfgeeks";
// Function calling
int ans1 = countSwap(s);
char[] charArray = s.ToCharArray();
Array.Reverse( charArray );
s = new string( charArray );
int ans2 = countSwap(s);
if(ans1 > ans2)
Console.WriteLine(ans1);
else
Console.WriteLine(ans2);
}
}
// This code is contributed by sapnasingh4991输出：
9
复杂性分析
时间复杂度：由于我们在字符串的长度上运行两个嵌套循环，时间复杂度为O(n 2 )
辅助空间：由于我们没有使用任何额外的空间，因此使用的辅助空间是O(1)
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