Visit original link: 704. Binary Search - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions for a better experience!

LeetCode link: 704. Binary Search, difficulty: Easy.

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.

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

Output: 4

Explanation: 9 exists in nums and its index is 4

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

Output: -1

Explanation: 2 does not exist in nums so return -1

• 1 <= nums.length <= 10000
• 104 < nums[i], target < 10000
• All the integers in nums are unique.
• nums is sorted in ascending order.

Because it is an already sorted array, by using the middle value for comparison, half of the numbers can be eliminated each time.

The fastest and easiest way is to use the three indices left, right, and middle. If nums[middle] > target, then right = middle - 1, otherwise, left = middle + 1.

• Time complexity: O(log N).
• Space complexity: O(1).

class Solution {
    public int search(int[] nums, int target) {
        var left = 0;
        var right = nums.length - 1;

        while (left <= right) {
            var middle = (left + right) / 2;
            
            if (nums[middle] == target) {
                return middle;
            }
            
            if (nums[middle] > target) {
                right = middle - 1;
            } else {
                left = middle + 1;
            }
        }

        return -1;
    }
}

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        left = 0
        right = len(nums) - 1

        while left <= right:
            middle = (left + right) // 2

            if nums[middle] == target:
                return middle

            if nums[middle] > target:
                right = middle - 1
            else:
                left = middle + 1

        return -1

class Solution {
public:
    int search(vector<int>& nums, int target) {
        auto left = 0;
        int right = nums.size() - 1; // Should not use 'auto' here because 'auto' will make this variable become `unsigned long` which has no `-1`.

        while (left <= right) {
            auto middle = (left + right) / 2;
            
            if (nums[middle] == target) {
                return middle;
            }
            
            if (nums[middle] > target) {
                right = middle - 1;
            } else {
                left = middle + 1;
            }
        }

        return -1;
    }
};

var search = function (nums, target) {
  let left = 0
  let right = nums.length - 1

  while (left <= right) {
    const middle = Math.floor((left + right) / 2)

    if (nums[middle] == target) {
      return middle
    }

    if (nums[middle] > target) {
      right = middle - 1
    } else {
      left = middle + 1
    }
  }

  return -1
};

public class Solution
{
    public int Search(int[] nums, int target)
    {
        int left = 0;
        int right = nums.Length - 1;

        while (left <= right)
        {
            int middle = (left + right) / 2;
            
            if (nums[middle] == target)
            {
                return middle;
            }
            
            if (nums[middle] > target)
            {
                right = middle - 1;
            } 
            else
            {
                left = middle + 1;
            }
        }

        return -1;
    }
}

func search(nums []int, target int) int {
    left := 0
    right := len(nums) - 1

    for left <= right {
        middle := (left + right) / 2

        if nums[middle] == target {
            return middle
        }

        if nums[middle] > target {
            right = middle - 1
        } else {
            left = middle + 1
        }
    }

    return -1
}

def search(nums, target)
  left = 0
  right = nums.size - 1

  while left <= right
    middle = (left + right) / 2

    return middle if nums[middle] == target

    if nums[middle] > target
      right = middle - 1
    else
      left = middle + 1
    end
  end

  -1
end

// Welcome to create a PR to complete the code of this language, thanks!

Dear LeetCoders! For a better LeetCode problem-solving experience, please visit website LeetCode.to: Dare to claim the best practices of LeetCode solutions! Will save you a lot of time!

Original link: 704. Binary Search - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions.

GitHub repository: leetcode-python-java.