class Solution {
    public int[] searchRange(int[] nums, int target) {

        return recur(0, nums.length - 1, nums, target);
    }

    public int[] recur(int left, int right, int[] nums, int target) {
        while (left <= right) {
            int mid = (left + right) >> 1;
            if (nums[mid] > target) {
                right = mid - 1;
            } else if (nums[mid] < target) {
                left = mid + 1;
            } else {
                if (nums[left] != target) {
		 // 因为left不是target需要向右移动
         因为nums{left}不等于target，需要向右移动去寻找，同理调用递归
                    left = recur(left + 1, mid, nums, target)[0];
                }
                if (nums[right] != target) {
                    right = recur(mid, right - 1, nums, target)[1];
                }
                return new int[] {left, right};
            }
        }
        return new int[] {-1, -1};
    }
}

思路：只要nums【mid】>=target的时候，说明还需要移动右端点，直到nums【mid】<target,然后左端点移动，导致left>right是循环结束，结果值就是上一次循环的结果。