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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0701_0800.s0704_binary_search;

// #Easy #Array #Binary_Search #Algorithm_I_Day_1_Binary_Search #Binary_Search_I_Day_1
// #Level_1_Day_7_Binary_Search #Udemy_Binary_Search
// #2022_03_23_Time_0_ms_(100.00%)_Space_54.8_MB_(20.10%)

/**
 * 704 - Binary Search\.
 *
 * 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.
 *
 * **Example 1:**
 *
 * **Input:** nums = [-1,0,3,5,9,12], target = 9
 *
 * **Output:** 4
 *
 * **Explanation:** 9 exists in nums and its index is 4
 *
 * **Example 2:**
 *
 * **Input:** nums = [-1,0,3,5,9,12], target = 2
 *
 * **Output:** -1
 *
 * **Explanation:** 2 does not exist in nums so return -1
 *
 * **Constraints:**
 *
 * *   1 <= nums.length <= 10⁴
 * *   -10⁴ < nums[i], target < 10⁴
 * *   All the integers in `nums` are **unique**.
 * *   `nums` is sorted in ascending order.
**/
public class Solution {
    public int search(int[] nums, int target) {
        int left = 0;
        int right = nums.length - 1;
        if (target < nums[left] || target > nums[right]) {
            return -1;
        }
        if (nums[left] == target) {
            return left;
        } else if (nums[right] == target) {
            return right;
        }
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (target == nums[mid]) {
                return mid;
            } else if (target > nums[mid]) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        return -1;
    }
}