LeetCode_Search in Rotated Sorted Array

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).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

找到左右分界点，然后判断target是在前半段还是后半段，然后在相应的地方用二分法查找

class Solution {
public:
 
bool binaryS(int A[], int min, int max, int &result, int target)
{
     if(A[min] > target || A[max] <target) {    
        result = -1 ;
        return false ;
     }
     
     int mid;
      
     while(min <= max)
     {
        mid = min + (max-min) /2 ;
        if(A[mid] == target) {
            result = mid;
            return true;
        }
        if(A[mid] < target)
            min = mid + 1;
         else
            max = mid -1; 
     }
     result = -1;
    
     return false;
}
    
int search(int A[], int n, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int i,result;
  
        if(n < 2)
        {
           for(i = 0; i<n ; i++)
             if(A[i] == target) break;
           return i<n ? i: -1;
        }
        
        
        int min = 0,max = n-1 ,mid;  
        while(min <= max)
        {
            mid = min + (max - min)/2 ;
            if(mid == 0 || mid == n-1) break;
           
            if(A[mid]<A[mid-1] && A[mid] <A[mid+1] )
                break;
             else if(A[mid] <A[0])
                    max = mid ;
                else if(A[mid] >A[0])
                    min =  mid + 1 ;
        }
        
        if(mid == 0)
        {
            if(A[0] > A[1])
               mid  = 1;
             else 
               mid = -1 ;
        }
        if(mid == n-1)
        {
            if(A[n-1] >A[n-2])
               mid = n ;      
        }
        
        if( mid == -1 || mid  == n )
           binaryS(A, 0,n - 1, result, target) ;
         else if(target >= A[0])
               binaryS(A, 0, mid - 1, result, target) ;
            else
               binaryS(A, mid, n-1, result, target);
               
        return result ;
    }
};

 另外一种思路：

class Solution {
public:
    int search(int A[], int n, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(A == NULL || n < 0) return -1;
        
        int left  = 0;
        int right = n-1;
        
        while(left <= right){
        
            int mid =  ((right - left)>>1) + left ;
            
            if(A[mid] == target)
                return mid;
            if(A[mid] > A[right]){
            
                if(target >= A[left] && target < A[mid])                
                    right = mid - 1;
                else 
                    left = mid  + 1;
            }else{
            
                if(target <= A[right] && target > A[mid])
                    left = mid + 1;
                else
                    right = mid - 1;
            }
        }
        
        return -1;
    }
};

 另一种思路：

class Solution {
public:
    int search(int A[], int n, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(A == NULL || n < 0) return -1;
        
        int left  = 0;
        int right = n-1;
        
        while(left <= right){
        
            int mid =  ((right - left)>>1) + left ;
            
            if(A[mid] == target)
                return mid;
            if(A[mid] > A[left]){
            
                if(target >= A[left] && target < A[mid])                
                    right = mid - 1;
                else 
                    left = mid  + 1;
            }else if(A[mid] < A[left]){
            
                if(target <= A[right] && target > A[mid])
                    left = mid + 1;
                else
                    right = mid - 1;
            }else{
                if(target == A[left])
                        return left;
                ++left;
            }
        }
        
        return -1;
    }
};