给定整数N ，任务是找到使N可以被60整除所需的最少交换次数。如果不可能，则打印“ -1 ”。
例子：
Input: N = 603
Output: 2
Explanation:
Two swap operations are required:
In the first swap (0, 3): 630
In the second swap (6, 3): 360
Now 360 is divisible by 60.
Therefore the minimum two swaps are required.
Input: N = 205
Output: -1
方法：
- 要将数字除以60，必须将其除以2、3和10。因此：
- 如果数字的数字总和不能被3整除，或者
- 如果没有可被2整除的数字，或者
- 如果该数字不包含任何0，
任何交换都不能使数字被60整除。因此，在这种情况下，需要-1交换。
- 然后，可以通过以下规则确定所需的最小交换次数：
- 如果该数字已经可以被60整除，则需要进行0次交换。可以确定最后一位数字(LSB)是否为0，而第二位最后一位数字是否可以被2整除。
- 如果以下任一情况为真，则需要进行1次交换。
- 如果最后一位数字(LSB)为0并且第二位最后一位数字不能被2整除，或者最后一位数字(LSB)被2分之一并且第二位最后一位数字为0。
- 如果最后一位数字(LSB)为0，而倒数第二位不能被2整除，则数字中存在多于1个零。
- 否则需要2次交换
下面是上述方法的实现
CPP
// C++ program to find minimum number
// of swap operations required
#include
using namespace std;
// Function that print minimum number
// of swap operations required
void MinimumSwapOperations(string s)
{
bool zero_exist = false;
bool multiple_of_2 = false;
int sum = 0;
int index_of_zero;
bool more_zero = false;
for (int i = 0; i < s.length(); i++) {
int val = s[i] - '0';
// Condition if more than one
// zero exist
if (zero_exist == true)
more_zero = true;
// Condition if zero_exist
if (val == 0) {
zero_exist = true;
index_of_zero = i;
}
// Computing total sum of all digits
sum += val;
}
// Condition if zero does not exist or
// the sum is not divisible by 3
if (zero_exist == false || sum % 3 != 0) {
cout << "-1"
<< "\n";
return;
}
for (int i = 0; i < s.length(); i++) {
int val = s[i] - '0';
// Condition to find a digit that is
// multiple of 2 other than one zero
if (val % 2 == 0 && i != index_of_zero)
multiple_of_2 = true;
}
// Condition if multiple of 2
// do not exist
if (multiple_of_2 == false) {
cout << "-1"
<< "\n";
return;
}
int last_val = s[s.length() - 1] - '0';
int second_last_val = s[s.length() - 2]
- '0';
// Condition for zero swaps
// means the number is already
// is divisible by 60
if (last_val == 0
&& second_last_val % 2 == 0)
cout << 0 << "\n";
// Condition for only one swap
else if ((last_val == 0
&& second_last_val % 2 != 0)
|| (last_val % 2 == 0
&& second_last_val == 0))
cout << 1 << "\n";
else if (more_zero == true
&& (last_val == 0
&& second_last_val % 2 != 0))
cout << 1 << "\n";
// Otherwise 2 swaps required
else
cout << 2 << "\n";
}
// Driver Code
int main()
{
string N = "20";
MinimumSwapOperations(N);
return 0;
} Java
// Java program to find minimum number
// of swap operations required
class GFG {
// Function that print minimum number
// of swap operations required
static void MinimumSwapOperations(String s)
{
boolean zero_exist = false;
boolean multiple_of_2 = false;
int sum = 0;
int index_of_zero = 0;
boolean more_zero = false;
for (int i = 0; i < s.length(); i++) {
int val = s.charAt(i) - '0';
// Condition if more than one
// zero exist
if (zero_exist == true)
more_zero = true;
// Condition if zero_exist
if (val == 0) {
zero_exist = true;
index_of_zero = i;
}
// Computing total sum of all digits
sum += val;
}
// Condition if zero does not exist or
// the sum is not divisible by 3
if (zero_exist == false || sum % 3 != 0) {
System.out.println("-1");
return;
}
for (int i = 0; i < s.length(); i++) {
int val = s.charAt(i) - '0';
// Condition to find a digit that is
// multiple of 2 other than one zero
if (val % 2 == 0 && i != index_of_zero)
multiple_of_2 = true;
}
// Condition if multiple of 2
// do not exist
if (multiple_of_2 == false) {
System.out.println("-1");
return;
}
int last_val = s.charAt(s.length() - 1)- '0';
int second_last_val = s.charAt(s.length() - 2)- '0';
// Condition for zero swaps
// means the number is already
// is divisible by 60
if (last_val == 0&& second_last_val % 2 == 0)
System.out.println(0);
// Condition for only one swap
else if ((last_val == 0
&& second_last_val % 2 != 0)
|| (last_val % 2 == 0
&& second_last_val == 0))
System.out.println(1);
else if (more_zero == true
&& (last_val == 0
&& second_last_val % 2 != 0))
System.out.println(1) ;
// Otherwise 2 swaps required
else
System.out.println(2) ;
}
// Driver Code
public static void main (String[] args)
{
String N = "20";
MinimumSwapOperations(N);
}
}
// This code is contributed by AnkitRai01Python3
# Python3 program to find minimum number
# of swap operations required
# Function that prminimum number
# of swap operations required
def MinimumSwapOperations(s):
zero_exist = False
multiple_of_2 = False
sum = 0
index_of_zero = 0
more_zero = False
for i in range(len(s)):
val = ord(s[i]) - ord('0')
# Condition if more than one
# zero exist
if (zero_exist == True):
more_zero = True
# Condition if zero_exist
if (val == 0):
zero_exist = True
index_of_zero = i
# Computing total sum of all digits
sum += val
# Condition if zero does not exist or
# the sum is not divisible by 3
if (zero_exist == False or sum % 3 != 0):
print("-1")
return
for i in range(len(s)):
val = ord(s[i]) - ord('0')
# Condition to find a digit that is
# multiple of 2 other than one zero
if (val % 2 == 0 and i != index_of_zero):
multiple_of_2 = True
# Condition if multiple of 2
# do not exist
if (multiple_of_2 == False):
print("-1")
return
last_val = ord(s[len(s) - 1]) - ord('0')
second_last_val = ord(s[len(s) - 2])- ord('0')
# Condition for zero swaps
# means the number is already
# is divisible by 60
if (last_val == 0
and second_last_val % 2 == 0):
print(0)
# Condition for only one swap
elif ((last_val == 0
and second_last_val % 2 != 0)
or (last_val % 2 == 0
and second_last_val == 0)):
print(1)
elif (more_zero == True
and (last_val == 0
and second_last_val % 2 != 0)):
print(1)
# Otherwise 2 swaps required
else:
print(2)
# Driver Code
if __name__ == '__main__':
N = "20"
MinimumSwapOperations(N)
# This code is contributed by mohit kumar 29C#
// C# program to find minimum number
// of swap operations required
using System;
class GFG {
// Function that print minimum number
// of swap operations required
static void MinimumSwapOperations(string s)
{
bool zero_exist = false;
bool multiple_of_2 = false;
int sum = 0;
int index_of_zero = 0;
bool more_zero = false;
for (int i = 0; i < s.Length; i++) {
int val = s[i] - '0';
// Condition if more than one
// zero exist
if (zero_exist == true)
more_zero = true;
// Condition if zero_exist
if (val == 0) {
zero_exist = true;
index_of_zero = i;
}
// Computing total sum of all digits
sum += val;
}
// Condition if zero does not exist or
// the sum is not divisible by 3
if (zero_exist == false || sum % 3 != 0) {
Console.WriteLine("-1");
return;
}
for (int i = 0; i < s.Length; i++) {
int val = s[i] - '0';
// Condition to find a digit that is
// multiple of 2 other than one zero
if (val % 2 == 0 && i != index_of_zero)
multiple_of_2 = true;
}
// Condition if multiple of 2
// do not exist
if (multiple_of_2 == false) {
Console.WriteLine("-1");
return;
}
int last_val = s[(s.Length - 1)- '0'];
int second_last_val = s[(s.Length - 2)- '0'];
// Condition for zero swaps
// means the number is already
// is divisible by 60
if (last_val == 0&& second_last_val % 2 == 0)
Console.WriteLine(0);
// Condition for only one swap
else if ((last_val == 0
&& second_last_val % 2 != 0)
|| (last_val % 2 == 0
&& second_last_val == 0))
Console.WriteLine(1);
else if (more_zero == true
&& (last_val == 0
&& second_last_val % 2 != 0))
Console.WriteLine(1) ;
// Otherwise 2 swaps required
else
Console.WriteLine(2) ;
}
// Driver Code
public static void Main (string[] args)
{
string N = "20";
MinimumSwapOperations(N);
}
}
// This code is contributed by AnkitRai01输出：
-1
时间复杂度： O(N)