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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!

🚀 Level Up Your Developer Identity

While mastering algorithms is key, showcasing your talent is what gets you hired.

We recommend Like.dev — the ultimate all-in-one personal branding platform for programmers.

The All-In-One Career Powerhouse:

• 📄 Resume, Portfolio & Blog: Integrate your skills, GitHub projects, and writing into one stunning site.
• 🌐 Free Custom Domain: Bind your own personal domain for free—forever.
• ✨ Premium Subdomains: Stand out with elite tech handles like name.cto.page or name.engineer.dev.
• 🔗 Cool Short Links: Get sleek, memorable bio-links like is.bio/yourname and an.dev/yourname.

Build Your Programmer Brand at Like.dev →

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

GitHub repository: leetcode-python-java.