how to find the closest 2 number in an array of unique positive number

One solution is to sort the array, then iterate through all neighboring pairs to find the pair with smallest difference. This is O(NlogN) solution. If the integers are bounded, you may be able to sort in O(N) time for an O(N) solution.

public static int[] find2Closest(int[] myArray) { assert(myArray.length >= 2): "myArray needs to be larger than 2"; Arrays.sort(myArray); //quickSort int indexMin1 = 0; int indexMin2 = 1; int min = myArray[indexMin1] - myArray[indexMin2]; for(int k = 0; k < myArray.length-1; k++) { int difference = Math.abs(myArray[k] - myArray[k+1]); if(difference < min) { min = difference; indexMin1 = k; indexMin2 = k+1; } } int[] values = new int[2]; values[0] = myArray[indexMin1]; values[1] = myArray[indexMin2]; return values; }

public class ClosestTwoNumber { public int[] findClosest2Number(int[] arr){ if(arr == null || arr.length diff) { min = diff; num1 = arr[i - 1]; num2 = arr[i]; } } if (num1 == -1 || num2 == -1) { throw new IllegalArgumentException(); } int[] result = new int[2]; result[0] = num1; result[1] = num2; return result; }

public class ClosestTwoNumber { public int[] findClosest2Number(int[] arr){ if(arr == null || arr.length diff) { min = diff; num1 = arr[i - 1]; num2 = arr[i]; } } if (num1 == -1 || num2 == -1) { throw new IllegalArgumentException(); } int[] result = new int[2]; result[0] = num1; result[1] = num2; return result; }