Next highest int from digits: Difference between revisions

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=={{header|Java}}==
=={{header|Java}}==
Additional testing is performed, including a number with all unique digits and a number with duplicate digits. Included test of all permutations, that the correct number of permutations is achieved, and that each number is greater than the previous number. If a library is not used, then this testing will provide a better proof of correctness.
Additional testing is performed, including a number with all unique digits and a number with duplicate digits. Included test of all permutations, that the order and correct number of permutations is achieved, and that each permutation is different than all others. If a library is not used, then this testing will provide a better proof of correctness.
<lang java>
<lang java>
import java.math.BigInteger;
import java.math.BigInteger;
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import java.util.ArrayList;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.List;
import java.util.Map;


public class NextHighestIntFromDigits {
public class NextHighestIntFromDigits {
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private static void testAll(String s) {
private static void testAll(String s) {
⚫
System.out.printf("Test all permutations of: %s%n", s);
⚫
int count = 1;
String sOrig = s;
⚫
System.out.printf("Test all iterations of: %s%n%d %s%n", s, count, format(s));
String sPrev = s;
String sPrev = s;
⚫
int count = 1;
// Check permutation order. Each is greater than the last
boolean orderOk = true;
Map <String,Integer> uniqueMap = new HashMap<>();
uniqueMap.put(s, 1);
while ( (s = next(s)).compareTo("0") != 0 ) {
while ( (s = next(s)).compareTo("0") != 0 ) {
count++;
count++;
System.out.printf("%d %s %s%n", count, format(s), Long.parseLong(s) > Long.parseLong(sPrev) ? "OK" : "NOT OK!!!");
if ( Long.parseLong(s) < Long.parseLong(sPrev) ) {
orderOk = false;
}
uniqueMap.merge(s, 1, (v1, v2) -> v1 + v2);
sPrev = s;
sPrev = s;
}
}
System.out.printf(" Order: OK = %b%n", orderOk);

// Test last permutation
String reverse = new StringBuilder(sOrig).reverse().toString();
System.out.printf(" Last permutation: Actual = %s, Expected = %s, OK = %b%n", sPrev, reverse, sPrev.compareTo(reverse) == 0);

// Check permutations unique
boolean unique = true;
for ( String key : uniqueMap.keySet() ) {
if ( uniqueMap.get(key) > 1 ) {
unique = false;
}
}
System.out.printf(" Permutations unique: OK = %b%n", unique);
// Check expected count.
Map<Character,Integer> charMap = new HashMap<>();
for ( char c : sOrig.toCharArray() ) {
charMap.merge(c, 1, (v1, v2) -> v1 + v2);
}
long permCount = factorial(sOrig.length());
for ( char c : charMap.keySet() ) {
permCount /= factorial(charMap.get(c));
}
System.out.printf(" Permutation count: Actual = %d, Expected = %d, OK = %b%n", count, permCount, count == permCount);

}
private static long factorial(long n) {
long fact = 1;
for (long num = 2 ; num <= n ; num++ ) {
fact *= num;
}
return fact;
}
}
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return sb.toString();
return sb.toString();
}
}

}
}
</lang>
</lang>
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9,589,776,899,767,587,796,600 -> 9,589,776,899,767,587,900,667
9,589,776,899,767,587,796,600 -> 9,589,776,899,767,587,900,667
3,345,333 -> 3,353,334
3,345,333 -> 3,353,334
Test all iterations of: 12345
Test all permutations of: 12345
Order: OK = true
1 12,345
Last permutation: Actual = 54321, Expected = 54321, OK = true
2 12,354 OK
Permutations unique: OK = true
3 12,435 OK
Permutation count: Actual = 120, Expected = 120, OK = true
4 12,453 OK
⚫
Test all permutations of: 11122
5 12,534 OK
Order: OK = true
... output removed to save space ...
Last permutation: Actual = 22111, Expected = 22111, OK = true
119 54,312 OK
Permutations unique: OK = true
120 54,321 OK
Permutation count: Actual = 10, Expected = 10, OK = true
⚫
Test all iterations of: 11122
1 11,122
2 11,212 OK
3 11,221 OK
4 12,112 OK
5 12,121 OK
6 12,211 OK
7 21,112 OK
8 21,121 OK
9 21,211 OK
10 22,111 OK
</pre>
</pre>


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