# 153 & 154. Find Minimum in Rotated Sorted Array I & II

**153.Find Minimum in Rotated Sorted Array**

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.`

**Example 1:**

<pre><code><strong>Input: nums = [3,4,5,1,2]
</strong><strong>Output: 1
</strong><strong>Explanation: The original array was [1,2,3,4,5] rotated 3 times.
</strong></code></pre>

**Example 2:**

<pre><code><strong>Input: nums = [4,5,6,7,0,1,2]
</strong><strong>Output: 0
</strong><strong>Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.
</strong></code></pre>

**Example 3:**

<pre><code><strong>Input: nums = [11,13,15,17]
</strong><strong>Output: 11
</strong><strong>Explanation: The original array was [11,13,15,17] and it was rotated 4 times. 
</strong></code></pre>

**Constraints:**

* `n == nums.length`
* `1 <= n <= 5000`
* `-5000 <= nums[i] <= 5000`
* All the integers of `nums` are **unique**.
* `nums` is sorted and rotated between `1` and `n` times.

***My Solutions:***

<pre class="language-java"><code class="lang-java">public int findMin(int[] nums) {
<strong>    int l = 0, r = nums.length - 1;
</strong>    while (l + 1 &#x3C; r) {
        int mid = l + (r - l) / 2;
        if (nums[mid] > nums[r]) l = mid;
        else r = mid;
    }
    return nums[l] > nums[r] ? nums[r] : nums[l];
}
</code></pre>

**154. Find Mininum in Rotated Sorted Array II**

和#153类似，但是会有重复数字。

***My Solutions:***

```java
public int findMin(int[] nums) {
    int l = 0, r = nums.length - 1;
    while (l + 1 < r) {
        int mid = l + (r - l) / 2;
        if (nums[mid] == nums[r]) r--; // 多加这一步。如果有重复数字，这样可以往前找
        else if (nums[mid] > nums[r]) l = mid;
        else r = mid;
    }
    if (nums[l] <= nums[r]) return nums[l];
    else return nums[r];
}
```