Suppose an array sorted in ascending order 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.
class Solution {
public:
int findMin(vector<int> &num) {
int start = ;
int end = num.size() - ;
int mid = ;
while (start < end) { //注意这里和普通的二分查找不同，这里是start < end不是 start <= end.
mid = start + (end - start)/;
if (num[mid] > num[end])
start = mid + ; //此时可以扔掉mid的值
else
end = mid;//此时不能扔掉mid的值
}
return num[end]; //退出循环说明start与end相等，所以只剩一个元素可能，所以return [start]或者return [end]都可以了。
//注意不能return mid，可以从{2,1}这个输入看出来。
}
};