【leetcode】33. Search rotation sort array

subject ：

33.  Search rotation sort array 
 An array of integers  nums  In ascending order , The values in the array   Different from each other  .
 Before passing it to a function ,nums  In some unknown subscript  k（0 <= k < nums.length） On the   rotate , Make array  [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]（ Subscript   from  0  Start   Count ）. for example , [0,1,2,4,5,6,7]  In subscript  3  It may turn into  [4,5,6,7,0,1,2] .

 Here you are.   After rotation   Array of  nums  And an integer  target , If  nums  There is a target value in  target , Returns its subscript , Otherwise return to  -1 .

 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
 Example  3：

 Input ：nums = [1], target = 0
 Output ：-1
 
 Tips ：
1 <= nums.length <= 5000
-10^4 <= nums[i] <= 10^4
nums  Each value in the   unique 
 Topic data assurance  nums  Rotated on a previously unknown subscript 
-10^4 <= target <= 10^4

Two points search ：

class Solution {
    
    public int search(int[] nums, int target) {
    
        int n = nums.length;
        int l = 0, r = n - 1;
        while (l <= r) {
    
            int mid = (l + r) / 2;
            if (nums[mid] == target){
    
                return mid;
            }
            //  The right half is in order ( Use the first and last arrays in the ordered half segment to judge whether the target value is within this region )
            else if (nums[r] > nums[mid]) {
    
                if(nums[mid] < target && nums[r] >= target ){
    
                    l = mid + 1;
                }
                else {
    
                    r = mid - 1;
                }
            }
            //  The left half is in order ( Use the first and last arrays in the ordered half segment to judge whether the target value is within this region )
            else {
    
                if(nums[mid] > target  && nums[l] <= target){
    
                    r = mid - 1;
                }
                else {
    
                    l = mid + 1;  
                }
            }   
        }
        return -1;
    }
}