Leetcode 0704. Binary Search

Given an array of integers `nums` which is sorted in ascending order, and an integer `target`, write a function to search `target` in `nums`. If `target` exists, then return its index. Otherwise, return `-1`. You must write an algorithm with `O(log n)` runtime complexity.

Easy

Description

Given an array of integers nums which is sorted in ascending order, and an integer target, write a function to search target in nums. If target exists, then return its index. Otherwise, return -1.

You must write an algorithm with O(log n) runtime complexity.

Example 1:

Input: nums = [-1,0,3,5,9,12], target = 9
Output: 4

• 9 exists in nums and its index is 4

Example 2:

Input: nums = [-1,0,3,5,9,12], target = 2
Output: -1

• 2 does not exist in nums so return -1

Constraints:

• 1 <= nums.length <= 10^4
• -10^4 < nums[i], target < 10^4
• All the integers in nums are unique.
• nums is sorted in ascending order.

Solution

class Solution {
  /**
   * Iterative a.k.a Binary Search
   * Time Complexity: BigO(logN)
   * Space Complexity: BigO(1)
   */
  public int search(int[] nums, int target) {
    int low = 0;
    int high = nums.length - 1;
    while (low <= high) {
      int mid = low + (high - low)/2;
      if (nums[mid] == target)
        return mid;
      else if (nums[mid] > target)
        high = mid - 1;
      else
        low = mid + 1;
    }
    return -1;
  }
}