[LeetCode]Find Minimum in Rotated Sorted Array II

Question

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.

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本题难度Hard。

二分查找法

【复杂度】
时间 O(N) 空间 O(N) 递归栈空间

【思路】
这道题难点就在于有重复数。与[LeetCode]Find Minimum in Rotated Sorted Array的思路差不多，其实只要增加对于nums[r]==nums[mid]的情况处理：让r向左移动，直到mid==r||nums[r]!=nums[mid]，然后再继续递归即可。中心思路就是抛弃多余的重复数（不是全部抛弃，有可能重复数就是最小值）。

【代码】

public class Solution {
    public int findMin(int[] nums) {
        return helper(0,nums.length-1,nums);
    }
    private int helper(int l,int r,int[] nums){
        //base case
        if(r==l)return nums[l];

        int mid=(l+r)/2;
        if(nums[r]>nums[mid])return helper(l,mid,nums);
        //addtional case
        if(nums[r]==nums[mid]){
            while(mid!=r&&nums[r]==nums[mid])r--;
            return helper(l,r,nums);
        }else
            return helper(mid+1,r,nums);
    }
}