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.

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 ;
}
};

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;
}
};

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;
}
};