第九周：[leetCode] 154. Find Minimum in Rotated Sorted Array II

Find Minimum in Rotated Sorted Array
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.
The array may contain duplicates.

解题分析：基于有序数列，可以利用二分查找定位最小元素，即mid = start + end。
根据递增顺序划分情况：
1、nums[end] < nums[mid]，最小值出现在mid-end之间，不包含mid；
2、nums[end] > nums[mid]，最小值出现在start-mid间；
3、nums[end] == nums[mid],:
(1) end-start-mid为重复值，最小值出现在mid-end之间;
(2) mid-end为重复值，最小值出现在start-mid之间；

代码实现：

    int findMin(vector<int>& nums) {        int start = 0, end = nums.size()-1;        while(start < end){            if(nums[start] < nums[end])return nums[start];            int mid = (start + end)/2;            if(nums[end] < nums[mid])                start = mid + 1;            else if(nums[end] > nums[mid])                end = mid;            else                 end--;        }        return nums[start];    }