# The Problem:

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.

# Notes:

- We could directly look up each element in the array which takes O(n) time, or we can take advantage of the sorted half and do binary search, each time examine if the target falling in the range of the sorted half, which should take O(logn) time.

# Java Solution

public class Solution { public int search(int[] A, int target) { if(A.length==0) return -1; int low = 0; int high = A.length-1; while(low <= high){ if(A[low]==target) return low; if(A[high]==target) return high; int mid = (low+high)/2; if(A[mid]==target) return mid; if(A[low]<A[mid]){ // former half sorted if(target > A[low] && target < A[mid]){ high = mid-1; } else { low = mid+1; } } else{ // the latter half sorted if(target > A[mid] && target < A[high]){ low = mid+1; } else { high = mid-1; } } } return -1; } }

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