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.

解题思路同Search in Rotated Sorted ArrayII。链接：http://blog.youkuaiyun.com/panzw2015/article/details/54317631

AC解：

public class Solution {
    private int ret=-1;
    public void setRet(int x)
    {
        ret=x;
    }
    
    public int search(int[] nums, int target) {
        int i;
        int nlen = nums.length;
        if(nlen==1)
        {
            if(nums[0]==target)
                return 0;
            else 
                return -1;
        }
        
        boolean flag = false; // not rotated
        for(i=1;i<nlen;i++)
        {
            if(nums[i]>=nums[i-1])
                continue;
            else 
            {
                flag = true; // rotated
                break;
            }
        }
        
        if(flag==false)
            find(nums,target,0,nlen-1);
        else
        {
            find(nums,target,0,i-1);
            find(nums,target,i,nlen-1);
        }
        
        return ret;
    }
    
    public void find(int[] nums,int target,int i,int j)
    {
        if(i<=j)
        {
            int l=i,r=j;
            int mid = l+(r-l)/2;
            if(nums[mid]==target)
                this.setRet(mid);
            else if(nums[mid]>target)
                find(nums,target,l,mid-1);
            else
                find(nums,target,mid+1,r);
        }
    }
}