# Share my java AC solution. O(log n) without recursion.

• Suppose we have three pointers:start, mid and end. There are three case for the array between start and end:

1. A[mid] < A[start]: Rotated, the maximum value is on the left of mid. If target < A[mid], then we search the left part; if target > A[mid], we then have two situations: (a) target <= A[end], we search the right part; (b) target > A[end], search the left part.

2. A[mid] > A[end]: Rotated, the maximum value is on the right of mid. target > A[mid]: search the left part; target < A[mid] && < A[start]: search the right part; target < A[mid] && >= A[start]: search the left part.

3. Otherwise: Not rotated. Just perform normal binary search.

My code:

``````public class Solution {
public int search(int[] A, int target) {
if (A.length==0) return -1;

int start = 0, end = A.length-1;
int mid = -1;

while (start<=end){
mid = (start+end)/2;
if (A[mid]==target) return mid;

if (A[mid]<A[start]){
if (target<A[mid])
end = mid-1;
else if (target<=A[end])
start = mid+1;
else end = mid-1;
continue;
}

if (A[mid]>A[end]){
if (target>A[mid]) start = mid+1;
else if (target>=A[start]) end = mid-1;
else start = mid+1;
continue;
}

if (target<A[mid]) end = mid-1;
else start = mid+1;
}

return -1;
}
``````

}

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