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Added search.js and solution.js
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‎easy/search.js

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//+ Jonas Raoni Soares Silva
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//@ http://raoni.org
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/**
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* @param {number[]} nums
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* @param {number} target
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* @return {number}
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*/
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var search = function(nums, target) {
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let h = nums.length;
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for (let l = -1, m; h - l > 1; nums[m = l + (h - l >>> 1)] < target ? l = m : h = m);
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return nums[h] !== target ? -1 : h;
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};

‎easy/search.md

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# [704. Binary Search](https://leetcode.com/problems/binary-search)
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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.
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You must write an algorithm with O(log n) runtime complexity.
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Example 1:
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Input: nums = [-1,0,3,5,9,12], target = 9
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Output: 4
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Explanation: 9 exists in nums and its index is 4
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Example 2:
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Input: nums = [-1,0,3,5,9,12], target = 2
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Output: -1
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Explanation: 2 does not exist in nums so return -1
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Constraints:
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1 <= nums.length <= 104
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-104 < nums[i], target < 104
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All the integers in nums are unique.
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nums is sorted in ascending order.

‎easy/solution.js

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//+ Jonas Raoni Soares Silva
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//@ http://raoni.org
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/**
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* Definition for isBadVersion()
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*
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* @param {integer} version number
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* @return {boolean} whether the version is bad
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* isBadVersion = function(version) {
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* ...
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* };
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*/
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/**
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* @param {function} isBadVersion()
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* @return {function}
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*/
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var solution = function(isBadVersion) {
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/**
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* @param {integer} n Total versions
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* @return {integer} The first bad version
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*/
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return function(h) {
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for (let l = -1, m; h - l > 1; isBadVersion(m = l + (h - l >>> 1)) ? h = m : l = m);
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return h;
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};
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};

‎easy/solution.md

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# [278. First Bad Version](https://leetcode.com/problems/first-bad-version)
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You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad version are also bad.
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Suppose you have n versions [1, 2, ..., n] and you want to find out the first bad one, which causes all the following ones to be bad.
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You are given an API bool isBadVersion(version) which returns whether version is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.
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Example 1:
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Input: n = 5, bad = 4
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Output: 4
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Explanation:
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call isBadVersion(3) -> false
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call isBadVersion(5) -> true
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call isBadVersion(4) -> true
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Then 4 is the first bad version.
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Example 2:
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Input: n = 1, bad = 1
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Output: 1
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Constraints:
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1 <= bad <= n <= 231 - 1

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