Find Minimum in Rotated Sorted Array

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题目：Find Minimum in Rotated Sorted Array

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.

解析：使用二分法找到最小元素，因为没有重复元素所以直接二分

代码：

class Solution {public:    int findMin(vector<int>& nums) {                        int left=0;        int right=nums.size()-1;        int mid=0;        if (nums[0]<=nums[nums.size()-1])        {           return nums[0];        }        while(left<right)        {           mid=(left+right)/2;         //  if (nums[mid])           if (nums[mid]>nums[nums.size()-1])           {               left=mid+1;           }           else if(nums[mid]<nums[nums.size()-1])           {               right=mid;           }                 }        return nums[left];            }};

Find Minimum in Rotated Sorted Array II

Follow up for "Find Minimum in Rotated Sorted Array":
What if duplicates are allowed?
Would this affect the run-time complexity? How and why?

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.

The array may contain duplicates.

解析：带有重复的元素，当mid的元素和左右两端的元素都相同时，无法判断下次二分时mid的位置应该是left,right,需要打破这种平衡，

然后把right--

代码：

class Solution {public:    int findMin(vector<int>& nums) {        int left=0;        int right=nums.size()-1;        int mid=0;        if (nums[0]<nums[nums.size()-1])        {           return nums[0];        }                while(left<right)        {           mid=(left+right)/2;         //  if (nums[mid])           if (nums[mid]>nums[right])           {               left=mid+1;           }           else if(nums[mid]<nums[right])           {               right=mid;           }           else if(nums[mid]==nums[right]&&nums[mid]==nums[left])           {               right--;           }           else if(nums[mid]==nums[right]&&nums[mid]!=nums[left])           {               right=mid;           }                           }        return nums[right];            }};

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