写一个方法判断数组内元素是否全部相同

function areAllElementsEqual(arr) {
  if (!arr || arr.length === 0) {
    return true; // Empty or null array is considered to have all elements equal
  }

  const firstElement = arr[0];

  for (let i = 1; i < arr.length; i++) {
    if (arr[i] !== firstElement) {
      return false;
    }
  }

  return true;
}


// Examples
console.log(areAllElementsEqual([1, 1, 1, 1])); // true
console.log(areAllElementsEqual([1, 2, 3, 4])); // false
console.log(areAllElementsEqual(['a', 'a', 'a'])); // true
console.log(areAllElementsEqual(['a', 'b', 'a'])); // false
console.log(areAllElementsEqual([])); // true - Empty array
console.log(areAllElementsEqual(null)); // true - Null array
console.log(areAllElementsEqual([NaN, NaN])); // false - NaN !== NaN
console.log(areAllElementsEqual([0, 0, 0, -0, 0])); // true - 0 and -0 are considered equal



// More robust version handling NaN (using every and isNaN):

function areAllElementsEqualIncludingNaN(arr) {
  if (!arr || arr.length === 0) {
    return true;
  }

  const firstElement = arr[0];

  return arr.every(element => {
    if (Number.isNaN(firstElement)) {
      return Number.isNaN(element);
    }
    return element === firstElement;
  });
}

console.log("Including NaN checks:");
console.log(areAllElementsEqualIncludingNaN([NaN, NaN])); // true
console.log(areAllElementsEqualIncludingNaN([NaN, 1]));   // false

This code provides two functions:

1. areAllElementsEqual(arr): This is a simple and efficient solution for most cases. It checks if all elements in an array are equal to the first element. It handles empty and null arrays gracefully. However, it doesn't handle NaN values correctly because NaN !== NaN.

2. areAllElementsEqualIncludingNaN(arr): This version uses the every() method and isNaN() to correctly handle NaN values. It's slightly less performant but more robust if you expect NaN values in your arrays.

Choose the function that best suits your needs based on whether you need to handle NaN correctly. The first function is generally sufficient and more efficient if you don't expect NaNs.