旋转数组

Posted 2022-07-16 shinered

假设按照升序排序的数组在预先未知的某个点上进行了旋转。

( 例如，数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。

搜索一个给定的目标值，如果数组中存在这个目标值，则返回它的索引，否则返回 -1 。

你可以假设数组中不存在重复的元素。

你的算法时间复杂度必须是 O(log n) 级别。

class Solution 
public:
    int search(vector<int>& nums, int target) 
        int left = 0;
        int right = nums.size()-1;
        int mid = (left+right)/2;
        if(nums.empty()) return -1;
        int result = -1;
         if(nums[mid] == target) return mid;
        while(left <= right)     
            //cout << mid << ",";
            if(nums[mid] == target)
                  result = mid;
                    break;
            
            
            if(nums[left] < nums[mid]) 
                if(nums[left] == target) return left;
                if(nums[mid] > target && nums[left] < target) 
                    right = mid-1;
                    mid = (right+left)/2;
                 else 
                    left = mid+1;
                    mid = (left+right) /2;                    
                                
             else 
                
                if(nums[right] == target) return right;
                if(nums[mid] < target && target <nums[right]  )
                    left = mid+1;
                    mid = (left+right) / 2;                    
                 else 
                    right = mid-1;
                    mid = (left+right) / 2;
                
                        
        
            
        return result;    
    
;