33. Search in Rotated Sorted Array

Suppose an array sorted in ascending order 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.

Your algorithm's runtime complexity must be in the order of _O_(log _n_).

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

属于有序数组的搜索，用二分搜索就好

public int search(int[] nums, int target) {
        int offset=0;
        for(int i=1;i<nums.length;i++){
            if(nums[i-1]>nums[i]){
                offset=i;
                break;
            }
        }
        int left=0;
        int right=nums.length-1;
        while(right>=left){
            int mid=(right+left)/2;
            int index=mid+offset;
            if(index>=nums.length) index-=nums.length;
            if(nums[index]==target) return index;
            else if(nums[index]<target) {
                left=mid+1;
            }else{
                right=mid-1;
            }
        }
        return -1;
}