[leedcode 154] 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.

public class Solution {
    public int findMin(int[] nums) {
        //注意：当mid等于left并且等于right时，需要使用常规方法进行寻找，其他情况和之前相同
        //注意单调递增的处理
        if(nums==null||nums.length<=0) return -1;
        int left=0;
        int right=nums.length-1;
        if(nums[left]<nums[right])return nums[left];/////
        while(left<right-1){
            int mid=(left+right)/2;
            if(nums[mid]==nums[left]&&nums[mid]==nums[right]){
                return find(nums,left,right);
            }
            if(nums[mid]>=nums[left]){
                left=mid;
            }else right=mid;
        }
        return nums[right];
    }
    public int find(int[] nums,int left,int right){
        int res=nums[left];
        for(int i=left+1;i<=right;i++){
            if(res>nums[i])
               res=nums[i];
        }
        return res;
    }
}