Classical Binary Search.java

/*
Find any position of a target number in a sorted array. Return -1 if target does not exist.

Example
Given [1, 2, 2, 4, 5, 5].
For target = 2, return 1 or 2.
For target = 5, return 4 or 5.
For target = 6, return -1.

time = O(log(n))
space = O(1)
*/

public class Solution {
  public int binarySearch(int[] array, int target) {
    if (array == null || array.length == 0) {
      return -1;
    }
    int left = 0;
    int right = array.length - 1;
    while (left <= right) {
      int mid = left + (right - left) / 2;
      if (array[mid] == target) {
        return mid;
      } else if (array[mid] > target) {
        right = mid - 1;
      } else {
        left = mid + 1;
      }
    }
    return -1;
  }
}