二进制搜索查找Magic Index

Question

I'm practicing binary search and I'm running into a wall when trying to implement it to find the "magic index" in an array. The magic index is A[i] == i .

I've found some implementations in Java using recursion , but I'm trying to avoid doing this recursively because recursion is expensive (and I want to see if binary search is appropriate here).

Here is my code:

function magicIndex(arr) {
  var result = arr, mid;
  while (result.length > 1) {
    mid = Math.floor(result.length / 2);

    if (result[mid] === mid) {
      return arr.indexOf(result[mid]);
    } else if (mid > result[mid]) {
      result = result.slice(mid+1, result.length);
    } else {
      result = result.slice(0, mid);
    }
  }
  return arr.indexOf(result.pop());
}

The problem is that the algorithm incorrectly slices the array to the wrong side on certain test runs.

For example, [-10, -3, 0, 2, 4, 8] returns 4 , but [-10, 1, 0, 2, 5, 8] returns 4 as well.

Answer 1

您的第二个数组未排序， 二进制搜索仅适用于已排序的数组。

Answer 2

Since you're practicing, I think it is good for you to code it by yourself but I'll give you a guide.

Use this algorithm (Note that input array should be sorted already, otherwise implement a sorting method):

  int arr[n];
  K ← 1; //search key
  l ← 0; r ← n - 1;
  while l ≤ r do
      m ← floor((l + r)/2)
      if K = arr[m]
          return m
      else if K < arr[m]
          r ← m − 1
      else
          l ← m + 1
  return −1

This will output the index of your search key, -1 if key is not found.