81. Search in Rotated Sorted Array II (Array; Divide-and-Conquer)

class Solution {
public:
    bool search(vector<int>& nums, int target) {
        return binarySearch(nums,0,nums.size()-1, target);
    }
    
    bool binarySearch(vector<int>& nums, int start, int end, int target){
        if(start==end){
            if(nums[start]==target) return true;
            else return false;
        }
        
        if(nums[start]==nums[end]) return binarySearch(nums,start,end-1,target); //ignore duplicate
        
        int mid = start+ ((end-start)>>1);
        //正序
        if(nums[mid]>=nums[start] && nums[mid]<nums[end]){ //mid可能=start，所以>=
            if(target <= nums[mid]) return binarySearch(nums,start,mid,target); //mid肯定<end，所以至少舍弃了一个
            else return binarySearch(nums,mid+1,end,target); //mid+1,至少舍弃了一个
        }
        
        //右侧rotate
        else if(nums[mid]>=nums[start] && nums[mid]>=nums[end]){
            if(target>=nums[start] && target<=nums[mid]) return binarySearch(nums,start,mid,target);
            else return binarySearch(nums,mid+1,end,target);
        }
        
        //左侧rotate
        else{
            if(target>=nums[start] || target<=nums[mid]) return binarySearch(nums,start,mid,target);
            else return binarySearch(nums,mid+1,end,target);
        }
    }
};