Sparse array search

I know a simple dichotomy algorithm, this time looking at the solution of the problem, I found a new continent.
It turns out that two points can still play like this.

class Solution {
    
    
    public int findString(String[] words, String s) {
    
    
        int left=0;
        int right=words.length-1;
        int temp=0;
        while(left<=right){
    
    
            while(left<=right&&words[left].equals("")) left++;
            while(left<=right&&words[right].equals("")) right--;
            //从两头向中间逼近。
            
            int mid=left+(right-left)/2;
            temp=mid;//记录mid的值，以方便后面使用
            while(words[mid].equals("")&&mid<right) mid=mid+1;
            //这里采用向后的遍历方式，找到不为空的字符串
            
            if(words[mid].equals("")){
    
    //到达了右边界，且为空，则直接调整右边界
                 right=mid-1;
                 continue;
            }
            if(words[mid].equals(s)) return mid;
            else if(words[mid].compareTo(s)<0)  left=mid+1;
            else right=mid-1;
        }
        return -1;
    }
}