题目： 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.
思路：①最原始的遍历一遍数组就好了；②二分查找，能减少算法的复杂度。
C++代码一（老老实实遍历一遍数组数据）：
class Solution{public:int findMin(vector<int> &num){for (size_t i = 1; i < num.size(); i++){if (num[i – 1] > num[i]) return num[i];}return num[0];}};
C++代码二（二分查找）：
class Solution{public:int findMin(vector<int> &num){size_t size = num.size() – 1;int left = 0;int right = int(size);int middle = right / 2;//注意这里循环条件有等号while(left <= right){middle = (left+ right) / 2;//如果middle比最后一个数字大，说明最小值在右边，反之，最小值在左边if (num[middle] > num[size]) left = middle + 1;else right = middle – 1;}return num[left];}};
C++代码三（对middle的求法和二不同）：
class Solution{public:int findMin(vector<int> &num){size_t size = num.size() – 1;int left = 0;int right = int(size);int middle = right / 2;while(left <= right){middle = left + (right – left) / 2;if (num[middle] > num[size]) left = middle + 1;else right = middle – 1;}return num[left];}};
C++代码四（right的求法和二不同，，注意while循环条件也会不同）：
class Solution{public:int findMin(vector<int> &num){size_t size = num.size() – 1;int left = 0;int right = int(size);int middle = right / 2;//这里循环条件不用等号while(left < right){middle = (right + left) / 2;if (num[middle] > num[size]) left = middle + 1;else right = middle;}return num[left];}};
C#代码：
public class Solution{public int FindMin(int[] num){int size = num.Length – 1;int left = 0;int right = size;int middle = right / 2;while (left < right){middle = (left + right) / 2;if (num[middle] > num[size]) left = middle + 1;else right = middle;}return num[right];}}
Python代码：
class Solution:# @param num, a list of integer# @return an integerdef findMin(self, num):size = len(num) – 1left = 0right = sizemiddle = right / 2while left < right:[middle] > num[size]:left = middle + 1else:[left]
我希望你能知道，我的心永远只为你跳动。
