11.5 sort the string array, where empty strings are scattered. Write a method to find the position of the given string.

The binary variation. When we find that the mid element is an empty string, we will also spread to both sides to find an index that is not a blank string as the mid, and continue to perform the second division.

 

public class Solution {    public int search(String[] list, String str) {        if (list == null || list.length == 0 || str == null || str.isEmpty())            return -1;        return searchHelper(list, str, 0, list.length - 1);    }    private int searchHelper(String[] a, String s, int left, int right) {        while (left <= right) {            int mid = left + (right - left) / 2;            if (a[mid].length() == 0) {                int leftNear = mid - 1;                int rightNear = mid + 1;                while (true) {                    if (leftNear < left && rightNear > right)                        return -1;                    if (leftNear >= left && a[leftNear].length() > 0) {                        mid = leftNear;                        break;                    }                    if (rightNear <= right && a[rightNear].length() > 0) {                        mid = rightNear;                        break;                    }                    leftNear--;                    rightNear++;                }            }            if (a[mid].equals(s))                return mid;            else if (a[mid].compareTo(s) < 0) {                left = mid + 1;            } else {                right = mid - 1;            }        }        return -1;    }    public static void main(String[] args) {        String[] stringList = { "apple", "", "", "banana", "", "", "", "carrot", "duck", "", "", "eel", "", "flower" };        System.out.println(new Solution().search(stringList, "carrot"));    }}

 

11.5 sort the string array, where empty strings are scattered. Write a method to find the position of the given string.