[LeetCode]Search in Rotated Sorted Array II

class Solution {
//classified discussion
//1. Based on the property of rotated array, there may or may not have one sorted sequence
//when one sequence is divided into two parts
//2. make decision under all these cases
public:
bool search(int A[], int n, int target) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
return BinarySearch(A, n, target);
}

bool BinarySearch( int* A, int n, int target )
{
//throw std::exception("The method or operation is not implemented.");
int l = 0;
int r = n-1;
while (l <= r)
{
int m = l+(r-l)/2;
if(A[m] == target) return true;
if (A[l] < A[m])//left subsequence sorted
{
if(A[l] <= target && target < A[m])
r = m-1;
else l = m+1;
}
else if (A[m] < A[r])
{
if(A[m] < target && target <= A[r])
l = m+1;
else r = m-1;
}
else if (A[l] == A[m])//A[m] is not the target, so remove en element equal to A[m] is safe
l++;
else if(A[m] == A[r])//ditto
r--;
}
return false;
}
};

class Solution {
public:
bool search(int A[], int n, int target) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int left = 0;
int right = n-1;
while(left <= right)
{
int mid = left+(right-left)/2;
if(A[left] < target && target < A[mid]) right = mid-1;
else if(A[mid] < target && target < A[right]) left = mid+1;
else
{
if(A[left] != target) left++;
else return true;
if(A[right] != target) right--;
else return true;
}
}
}
};