## 704. Binary Search
之前面試有考刷題，所以有刷過一陣子題目（~~不過都是Easy就是了~~），所以想紀錄之前紀錄一些基礎題。

>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

### 需要已排序陣列中，尋找到 target 數字

#### 思路
* 將數列從中間劃分為兩半，根據目標值與中間值的比較來決定搜尋哪一半。
這個過程會反覆縮小搜尋範圍，直到找到目標或搜尋範圍為空。
時間複雜度：`O(log n)`，因為每次都將搜尋範圍減半。

> 其實簡單來說就是小時候玩的終極密碼

### Code

function search(nums, target) {
    let low = 0;
    let high = nums.length - 1;

    while (low <= high) {
        let mid = Math.floor((low + high) / 2);
        if (nums[mid] === target) {
            return mid;
        }
        if (nums[mid] < target) {
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }

    return -1;
}

const nums1 = [-1, 0, 3, 5, 9, 12];
const target1 = 9;

console.log(search(nums1, target1)); // 4