When there is a pivot, you take a mid, and two parts are obtained: one sorted part and the other half. Binary search inside the sorted part.

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class Solution {
    public int search(int[] nums, int target) {
        if (nums.length == 0) return -1;
        int left = 0, right = nums.length - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) return mid;
          
            else if (nums[mid] < nums[right]) {
                if (nums[mid] < target && target <= nums[right]) left = mid+1;
                else right = mid-1;
            } else {
                if (nums[mid] > target && nums[left] <= target) right = mid-1;
                else left = mid + 1;
            }
        }
        return -1;
    }
}

nums[mid] < nums[right]: a prev comparison before formal nums[mid] < target and nums[mid] > target