如果你保证数组是有序的，没有任何重复，那么你可以在O（1）中检查它
我认为这段代码应该能在这个特定的情况下工作：）

//assume that given array is ordered and has no duplicated value
long[] myarray = {5,6,7};                 //no gap
long[] myarray1 = {1,2,4};                //has gap
long[] myarray2 = {10,11,12,13,14,15};    //no gap

//return true if has gap
//return false if no gap
//throw null-pointer if empty
public static boolean checkIfHasGap(long[] array) {
    if (array.length == 0) {
        throw new NullPointerException("Given Array is empty");
    } else {
        return array[0] + array.length != array[array.length - 1] + 1;
    }
}

public static boolean hasGaps(long[] array) {
    if (array == null || array.length == 0) {
        return false;
    }
    if (array[array.length - 1] - array[0] + 1 != array.length) {
        return true;
    }
    for (int i = 1; i < array.length; i++) {
        if (array[i] != array[i - 1] + 1) {
            return true;
        }
    }
    return false;
}

此方法首先检查“简单”的情况，然后迭代数组以检查更复杂的情况。

遍历数组并检查当前项是否正好比当前索引大x，其中x是数组中的第一个元素。

public boolean hasGaps(long[] array) {
    if(array == null || array.length == 0) {
        return false;
    }

    long start = array[0];
    for(int i = 0; i < array.length; i++) {
        if(array[i] != i + start) {
            return true;
        }
    }
    return false;
}

更简单可靠的方法：

const getMissingNumbers = (list) => {
  if (!list || list.length === 0) return [];

  const missing = [];
  // sort the list
  list.sort((a, b) => a - b);
  // pin start and end points in the list
  const step = list[0];
  const last = list[list.length - 1];
  for (let i = step; i < last; i++) {
    if (list.indexOf(i) === -1) missing.push(i);
  }
  return missing;
}