给定一个字符串，找到最少数量的交换(不一定相邻)以将其转换为并排具有相似字符的字符串。
例子：
Input : abcb
Output : 1
Explanation : swap (c, b) to form abbc or acbb. Number of swap operations for this is 1;
Input : abbaacb
0123456
Output : 2
Explanation : Swap 0th index with 6th index and then swap 5th index with 6th index.
这个想法是要考虑由两个元素之间的每次交换形成的所有排列，并且也不要交换两个元素。
C++
#include
using namespace std;
// checks whether a string has similar characters side by side
bool sameCharAdj(string str)
{
int n = str.length(), i;
set st;
st.insert(str[0]);
for (i = 1; i < n; i++) {
// If similar chars side by side, continue
if (str[i] == str[i - 1])
continue;
// If we have found a char equal to current
// char and does not exist side to it,
// return false
if (st.find(str[i]) != st.end())
return false;
st.insert(str[i]);
}
return true;
}
// counts min swap operations to convert a string
// that has similar characters side by side
int minSwaps(string str, int l, int r, int cnt, int minm)
{
// Base case
if (l == r) {
if (sameCharAdj(str))
return cnt;
else
return INT_MAX;
}
for (int i = l + 1; i <= r; i++) {
swap(str[i], str[l]);
cnt++;
// considering swapping of i and l chars
int x = minSwaps(str, l + 1, r, cnt, minm);
// Backtrack
swap(str[i], str[l]);
cnt--;
// not considering swapping of i and l chars
int y = minSwaps(str, l + 1, r, cnt, minm);
// taking min of above two
minm = min(minm, min(x, y));
}
return minm;
}
// Driver code
int main()
{
string str = "abbaacb";
int n = str.length(), cnt = 0, minm = INT_MAX;
cout << minSwaps(str, 0, n - 1, cnt, minm) << endl;
return 0;
} Java
import java.util.*;
class GFG {
// checks whether a String has similar characters side by side
static boolean sameJavaharAdj(char str[]) {
int n = str.length, i;
TreeSet st = new TreeSet<>();
st.add(str[0]);
for (i = 1; i < n; i++) {
// If similar chars side by side, continue
if (str[i] == str[i - 1]) {
continue;
}
// If we have found a char equal to current
// char and does not exist side to it,
// return false
if (st.contains(str[i]) & (str[i] != st.last())) {
return false;
}
st.add(str[i]);
}
return true;
}
// counts min swap operations to convert a String
// that has similar characters side by side
static int minSwaps(char str[], int l, int r, int cnt, int minm) {
// Base case
if (l == r) {
if (sameJavaharAdj(str)) {
return cnt;
} else {
return Integer.MAX_VALUE;
}
}
for (int i = l + 1; i <= r; i++) {
swap(str, i, l);
cnt++;
// considering swapping of i and l chars
int x = minSwaps(str, l + 1, r, cnt, minm);
// Backtrack
swap(str, i, l);
cnt--;
// not considering swapping of i and l chars
int y = minSwaps(str, l + 1, r, cnt, minm);
// taking Math.min of above two
minm = Math.min(minm, Math.min(x, y));
}
return minm;
}
static void swap(char[] arr, int i, int j) {
char temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
// Driver code
public static void main(String[] args) {
String str = "abbaacb";
int n = str.length(), cnt = 0, minm = Integer.MAX_VALUE;
System.out.print(minSwaps(str.toCharArray(), 0, n - 1, cnt, minm));;
}
}
// This code is contributed Rajput-Ji Python3
# Python3 implementation of the approach
from sys import maxsize
# checks whether a string has
# similar characters side by side
def sameCharAdj(string):
n = len(string)
st = set()
st.add(string[0])
for i in range(1, n):
# If similar chars side by side, continue
if string[i] == string[i - 1]:
continue
# If we have found a char equal to current
# char and does not exist side to it,
# return false
if string[i] in st:
return False
st.add(string[i])
return True
# counts min swap operations to convert a string
# that has similar characters side by side
def minSwaps(string, l, r, cnt, minm):
# Base case
if l == r:
if sameCharAdj(string):
return cnt
else:
return maxsize
for i in range(l + 1, r + 1, 1):
string[i], string[l] = string[l], string[i]
cnt += 1
# considering swapping of i and l chars
x = minSwaps(string, l + 1, r, cnt, minm)
# Backtrack
string[i], string[l] = string[l], string[i]
cnt -= 1
# not considering swapping of i and l chars
y = minSwaps(string, l + 1, r, cnt, minm)
# taking min of above two
minm = min(minm, min(x, y))
return minm
# Driver Code
if __name__ == "__main__":
string = "abbaacb"
string = list(string)
n = len(string)
cnt = 0
minm = maxsize
print(minSwaps(string, 0, n - 1, cnt, minm))
# This code is contributed by
# sanjeev2552C#
using System;
using System.Collections.Generic;
class GFG
{
// checks whether a String has similar
// characters side by side
static bool sameJavaharAdj(char []str)
{
int n = str.Length, i;
HashSet st = new HashSet();
st.Add(str[0]);
for (i = 1; i < n; i++)
{
// If similar chars side by side, continue
if (str[i] == str[i - 1])
{
continue;
}
// If we have found a char equal to current
// char and does not exist side to it,
// return false
if (st.Contains(str[i]) & !st.Equals(str[i]))
{
return false;
}
st.Add(str[i]);
}
return true;
}
// counts min swap operations to convert a String
// that has similar characters side by side
static int minSwaps(char []str, int l, int r,
int cnt, int minm)
{
// Base case
if (l == r)
{
if (sameJavaharAdj(str))
{
return cnt;
}
else
{
return int.MaxValue;
}
}
for (int i = l + 1; i <= r; i++)
{
swap(str, i, l);
cnt++;
// considering swapping of i and l chars
int x = minSwaps(str, l + 1, r, cnt, minm);
// Backtrack
swap(str, i, l);
cnt--;
// not considering swapping of i and l chars
int y = minSwaps(str, l + 1, r, cnt, minm);
// taking Math.min of above two
minm = Math.Min(minm, Math.Min(x, y));
}
return minm;
}
static void swap(char[] arr, int i, int j)
{
char temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
// Driver code
public static void Main()
{
String str = "abbaacb";
int n = str.Length, cnt = 0, minm = int.MaxValue;
Console.WriteLine(minSwaps(str.ToCharArray(), 0,
n - 1, cnt, minm));;
}
}
// This code is contributed mits 输出：
2
时间复杂度：递归是T(n)= 2n * T(n-1)+ O(n)
因此时间复杂度大于O((2 * n)！)