Suppose a sorted array is rotated at some pivot unknown to you beforehand.
(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).
Find the minimum element.
You may assume no duplicate exists in the array.

O(N)时间复杂度解法

无论数组如何旋转，都能保证：如果下一个元素开始变小，那么下一个元素将是最小的。如果一直未变小，那么首元素将是最小的。

C++ code
    int findMin(vector<int> &num) {
        for (int i = 1; i < num.size(); ++i)
            if (num[i] < num[i - 1])
                return num[i];

        return num[0];
    }

O(log(N))时间复杂度解法

Discuss上看到一个利用二分查找的更好的方法，时间复杂度为O(log(N))。

Looking at subarray with index [start,end]. We can find out that if the first member is less than the last member, there‘s no rotation in the array. So we could directly return the first element in this subarray.
If the first element is larger than the last one, then we compute the element in the middle, and compare it with the first element. If value of the element in the middle is larger than the first element, we know the rotation is at the second half of this array. Else, it is in the first half in the array.

C++ code
int findMin(vector<int> &num) {
        int start=0,end=num.size()-1;

        while (start<end) {
            if (num[start]<num[end])
                return num[start];

            int mid = (start+end)/2;

            if (num[mid]>=num[start]) {
                start = mid+1;
            } else {
                end = mid;
            }
        }

        return num[start];
    }

LeetCode[Array]: Find Minimum in Rotated Sorted Array