You'll need a helper variable for this, so depending on your programming language, the solution becomes:
int helper;
helper = arr[i];
arr[i] = arr[j];
arr[j] = helper;
Answer:
The solution code is written in Python 3:
- def modifyList(listNumber):
- posCount = 0
- negCount = 0
-
- for x in listNumber:
- if x > 0:
- posCount += 1
- else:
- negCount += 1
-
- if(posCount == len(listNumber)):
- listNumber.append(max(listNumber))
-
- if(negCount == len(listNumber)):
- listNumber.append(min(listNumber))
-
- print(listNumber)
-
- modifyList([-1,-99,-81])
- modifyList([1,99,8])
- modifyList([-1,99,-81])
Explanation:
The key step to solve this problem is to define two variables, posCount and negCount, to track the number of positive value and negative value from the input list (Line 2 - 3).
To track the posCount and negCount, we can traverse through the for-loop and create if else statement to check if the current number x is bigger than 0 then increment posCount by 1 otherwise increment negCount (Line 5- 9).
If all number in the list are positive, the posCount should be equal to the length of the input list and the same rule is applied to negCount. If one of them happens, the listNumber will append either the maximum number (Line 11 -12) or append the minimum number (Line 14-15).
If both posCount and negCount are not equal to the list length, the block of code Line 11 -15 will be skipped.
At last we can print the listNumber (Line 17).
If we test our function using the three sets of input list, we shall get the following results:
[-1, -99, -81, -99]
[1, 99, 8, 99]
[-1, 99, -81]
Answer:
1010 0111 1101 1111
Explanation:
A = 10 in decimal = 1010 in binary
7 = 7 in decimal = 0111 in binary
D = 13 in decimal = 1101 in binary
F = 15 in decimal = 1111 in binary
Therefore 0xA7DF = 1010 0111 1101 1111 in binary
```
#!/usr/local/bin/python3
foo = float( input( "Enter a number: " ) )
if( foo < 0.0 ):
print( "negative" )
elif( foo > 0.0 ):
print( "positive" )
else
print( "zero" )
exit( 0 )
```
