# 153. Find Minimum in Rotated Sorted Array

### Leetcode

<https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/>

### 題目

Suppose an array of length `n` sorted in ascending order is **rotated** between `1` and `n` times. For example, the array `nums = [0,1,2,4,5,6,7]` might become:

* `[4,5,6,7,0,1,2]` if it was rotated `4` times.
* `[0,1,2,4,5,6,7]` if it was rotated `7` times.

Notice that **rotating** an array `[a[0], a[1], a[2], ..., a[n-1]]` 1 time results in the array `[a[n-1], a[0], a[1], a[2], ..., a[n-2]]`.

Given the sorted rotated array `nums` of **unique** elements, return *the minimum element of this array*.

You must write an algorithm that runs in `O(log n) time.`

&#x20;

**Example 1:**

```
Input: nums = [3,4,5,1,2]
Output: 1
Explanation: The original array was [1,2,3,4,5] rotated 3 times.
```

**Example 2:**

```
Input: nums = [4,5,6,7,0,1,2]
Output: 0
Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.
```

**Example 3:**

```
Input: nums = [11,13,15,17]
Output: 11
Explanation: The original array was [11,13,15,17] and it was rotated 4 times. 
```

### 解答 <a href="#ti-jie" id="ti-jie"></a>

* **方法一**

如果不限制用 `O(log N)` 解的話，只要遍歷矩陣找出下一個元素大於目前元素時，就是矩陣旋轉的地方

```javascript
var findMin = function (nums) {
    for (let i = 0; i < nums.length; i++) {
        if (nums[i + 1] < nums[i]) {
            return nums[i + 1];
        }
    }

    return nums[0];
};
```

空間複雜度： O(n)

* **方法二**

由於矩陣在旋轉點前、後都是由小到大排序，所以最快可以用二分搜尋法達到 `O(lon N)` 的時間複雜度

```javascript
var findMin = function (nums) {
    let low = 0;
    let high = nums.length - 1;

    while (low <= high) {
        const mid = Math.floor((low + high) / 2);
        // 找到旋轉點
        if (nums[mid + 1] < nums[mid]) {
            return nums[mid + 1];
        } else if (nums[mid] >= nums[0]) {
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }

    return nums[0];
};
```
