[leetcode]Find Minimum in Rotated Sorted Array II

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 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.

The array may contain duplicates.

class Solution {public:    int findMin(vector<int>& nums, int b, int e){        int minVal = nums[b];        for(int i = b + 1; i <= e; i++){            minVal = min(minVal, nums[i]);        }        return minVal;    }    int findMin(vector<int>& nums) {        int mid;          int left = 0;          int right = nums.size() - 1;                  while(left < right){              mid = left + (right - left) / 2;            if(nums[mid] == nums[right] && nums[mid] == nums[left]){                 return findMin(nums, left, right);            }            if(nums[mid] == nums[left]){                if(nums[mid] < nums[right]){ // --/-                    return findMin(nums, mid, right);                }                  if(nums[mid] > nums[right]){ //--\ -                    left = mid + 1;                    continue;                }            }            if(nums[mid] == nums[right]){                if(nums[mid] < nums[left]){// \--                      return findMin(nums, left, mid);                }                if(nums[mid] > nums[left]){//  /--   -                    right = mid - 1;                    continue;                }            }                        if(nums[mid] > nums[right]){ //  1、/\  2、/                  left = mid + 1;              }else{                  //正好位于拐点 3、\/ 4、\  -                  if(nums[mid] < nums[mid - 1]){                      return nums[mid];                  }                  right = mid - 1;              }          }                    return nums[left];             }};