# My straight forward Java solution with explanation

• The idea is straightforward
• since only one rotation is made, there are two sorted arrays
• find the pivot, and decide to search the target in which array
• binary search then

Attached is the accepted code

``````public class Solution {
public int search(int[] nums, int target) {
int n = nums.length;
if(n==0 || (n==1 && nums[0]!=target)) return -1;
if(n==1 && nums[0]==target) return 0;
int pivot=0,i,lo,hi;
for(i=0;i<n-1;i++)
if(nums[i]>nums[i+1]) break;
pivot = i;
if(pivot==n-1) return bsearch(0,n-1,target,nums);
return target>=nums[0]?bsearch(0,pivot,target,nums):bsearch(pivot+1, n-1,target,nums);
}
private int bsearch(int a, int b, int target, int[] nums){
if(target>nums[b]||target<nums[a]) return -1;
while(a<=b){
int mid = (a+b)/2;
if(target==nums[mid]) return mid;
else if(target>nums[mid]) a = mid+1;
else b = mid-1;
}
return -1;
}
}
``````
• AC, 0ms

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