feat(problems): add binary search solution for LeetCode 704

work/Easy/Array/704.binary-search.js

@@ -0,0 +1,61 @@
+/*
+ * 704. Binary Search
+ * Difficulty: Easy
+ * https://leetcode.com/problems/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.
+ * 
+ *  
+ * 
+ * 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 <= 10^4
+ * 
+ * 	• -10^4 < nums[i], target < 10^4
+ * 
+ * 	• All the integers in nums are unique.
+ * 
+ * 	• nums is sorted in ascending order.
+*/
+
+/**
+ * @param {number[]} nums
+ * @param {number} target
+ * @return {number}
+ */
+var search = function(nums, target) {
+    let left = 0;
+    let right = nums.length - 1;
+
+    while(left <= right) {
+        const mid = left + Math.floor((right - left) / 2);
+        if (nums[mid] === target) {
+            return mid;
+        } else if(nums[mid] < target){
+            left = mid + 1;
+        } else if (nums[mid] > target) {
+            right = mid - 1;
+        }
+    }
+    return -1;
+};