Use the Arrays.sort() method, the lowest value will be element0.
In terms of performance, this should not be expensive since the sort operation is already optimised. Also has the advantage of being concise.
private int min(int ... value) {
Arrays.sort(value);
return value[0];
}
Proof of concept
int[] intArr = {12, 5, 6, 9, 44, 28, 1, 4, 18, 2, 66, 13, 1, 33, 74, 12,
5, 6, 9, 44, 28, 1, 4, 18, 2, 66, 13};
// Sorting approach
long startTime = System.currentTimeMillis();
int minVal = min(intArr);
long endTime = System.currentTimeMillis();
System.out.println("Sorting: Min => " + minVal + " took => " + (endTime -
startTime));
System.out.println(startTime + " " + endTime);
System.out.println(" ");
// Scanning approach
minVal = 100;
startTime = System.currentTimeMillis();
for(int val : intArr) {
if (val < minVal)
minVal = val;
}
endTime = System.currentTimeMillis();
System.out.println("Iterating: Min => " + minVal + " took => " + (endTime
- startTime));
System.out.println(startTime + " " + endTime);
