Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value.

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

If the target is not found in the array, return [-1, -1].

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

思路：就是binary search，一個search左邊一個search 右邊；注意邊界情況，一次完成，開心！

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int[] res = {-1,-1};
        if(nums == null || nums.length == 0) return res;
        res[0] = findleft(nums, target, 0, nums.length-1, true);
        res[1] = findleft(nums, target, 0, nums.length-1, false);
        return res;
    }
    
    public int findleft(int[] nums, int target, int start, int end, boolean findleft){
        if(start > end){
            return -1;
        } else {
            int mid = start+(end-start)/2;
            if(findleft){
                 if(nums[mid] == target){
                    if((mid-1>=0 && nums[mid-1] != target) || mid == 0){
                          return mid;
                      } else {
                         return findleft(nums, target, start, mid-1, findleft);
                      }
                } else if(nums[mid] > target){
                     return findleft(nums, target, start, mid-1, findleft);
                } else { // nums[mid] < target;
                     return findleft(nums, target, mid+1, end, findleft);
                 }
            } else { // find right index;
                if(nums[mid] == target){
                    if((mid+1<=nums.length-1 && nums[mid+1] != target) || mid == nums.length-1){
                          return mid;
                      } else {
                         return findleft(nums, target, mid+1, end, findleft);
                      }
                } else if(nums[mid] > target){
                     return findleft(nums, target, start, mid-1, findleft);
                } else { // nums[mid] < target;
                     return findleft(nums, target, mid+1, end, findleft);
                 }
            }
        }
    }
}