Thought:
Binary Search with different condition.
Code:
public class Solution {
public int search(int[] A, int target) {
int start = 0;
int end = A.length - 1;
while (start <= end) {
int mid = (start + end) / 2;
if (A[mid] == target) return mid;
if (A[start] > A[mid]) {
if (A[mid] <= target && target <= A[end]) start = mid + 1;
else end = mid - 1;
}else {
if (A[start] <= target && target <= A[mid]) end = mid - 1;
else start = mid + 1;
}
}
return - 1;
}
}
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