154. Find Minimum in Rotated Sorted Array II (Array; Divide-and-Conquer)

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.

 

思路：有重复元素的时候，不能按I中这么分三类（反例：Input: [3,3,1,3]=>将roate in right判断成没有rotate）。解决方法是，当碰到nums[start]=nums[end]的情况时，end-1，寻找不同元素再进行二分法。

class Solution {
public:
    int findMin(vector<int>& nums) {
        return dfs(nums,0,nums.size()-1);
    }
    
    int dfs(vector<int>& nums, int start, int end){
        if(start==end) return nums[start];
        if(nums[start]==nums[end]) return dfs(nums,start, end-1);
        
        int mid = start + ((end - start) >> 1);
        if(nums[mid] > nums[end]){ //rotate in the right
            return dfs(nums, mid+1,end);
        }
        else if(nums[start] <= nums[mid]){ //no rotate
            return nums[start];
        }
        else{ //rorate in the left
            return dfs(nums, start, mid);
        }
    }
};