33. Search in Rotated Sorted Array

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.</br> (i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).</br> You are given a target value to search. If found in the array return its index, otherwise return -1.

class Solution {
    public int search(int[] nums, int target) {
        int lo = 0;
        int hi = nums.length-1;
        // find the piviot
        // binary search without = 
        // then lo = mid+1;
        while(lo<hi){
            int mid = (lo+hi)/2;
            if(nums[mid]>nums[hi]){
                lo=mid+1;
            }else{
                hi=mid;
            }
        }
        int piviot = lo;
        lo = 0;
        hi = nums.length-1;
        /*
        binary search with =
        then lo = mid+1; hi = mid-1
        this method pretend to do binary search in unrotated
        When it comes to compare, it find the value from 
        rotated list.
        */
        while(lo<=hi){
            int mid = (lo+hi)/2;
            int realmid = (mid+piviot)%nums.length;
            if(nums[realmid]==target) return realmid;
            if(nums[realmid]<target) lo = mid+1;
            else hi = mid-1;
        }
        return -1;
    }
}