java - How to find mode of an array, if one exists. If more than one mode, return NaN -

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• write mode method in java find occurring element in array 13 answers

as title states, need find mode of array. however, there few stipulations this:

1) if no mode exists (i.e. each element appears once, or equal times) return double.nan

2) if more 1 mode exists (i.e. {1,2,2,3,4,5,5,6} give 2 modes, 2 , 5) return double.nan

basically, should return element of array if mode of array, , appears @ least once more other elements. other time, should return double.nan

my current code returns mode. however, if 2 numbers appear equally, returns latter of 2 mode, not nan. also, doesn't return nan if no mode exists.

any appreciated.

here have far:

public double mode(){     double[] holder = new double[data.length];     double tempmax = 0, permmax = 0, location = 0, arraymode = 0;      (int = 0; < data.length; ++i) {         int count = 0;         (int j = 0; j < data.length; ++j) {             if (data[j] == data[i])                 ++count;         }         holder[i] = count;     }      (int w = 0; w < holder.length; w++){     if (holder[w] > tempmax){         tempmax = holder[w];         arraymode = data[w];     }     }     permmax = arraymode;      return permmax; } 

it easy find duplicate question, e.g. here.

that being said, i'd add solution:

public double findmode(int[] inarray) {         list<integer> il = intstream.of(inarray).boxed().collect(collectors.tolist());         int maxfreq = 0;         double value = 0;         (integer : ia){             if (collections.frequency(il, i) > maxfreq && != value){                 maxfreq = collections.frequency(il, i);                 value = i;             } else if (collections.frequency(il, i) == maxfreq && != value) {                 value = double.nan;              }         }         return value; } 

it turns int[] list<integer> , uses collections.frequency method number of occurrences of each value.

disclaimer: there optimizations missed.