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;

    }

};