Sorting&Searching 二分法找有空字符串的数组 @CareerCup

要点是找到最近的不是空字符串的位置，然后作为mid，再继续二分

package Sorting_Searching;/** * Given a sorted array of strings which is interspersed with empty strings, * write a method to find the location of a given string. *  * Example: find “ball” in [“at”, “”, “”, “”, “ball”, “”, “”, “car”, “”,“”, * “dad”, “”, “”] will return 4 *  * Example: find “ballcar” in [“at”, “”, “”, “”, “”, “ball”, “car”, “”, “”, * “dad”, “”, “”] will return -1 *  * 译文： *  * 给你一个排好序的并且穿插有空字符串的字符串数组，写一个函数找到给定字符串的位置。 *  * 例子：在字符串数组 [“at”, “”, “”, “”, “ball”, “”, “”, “car”, “”,“”, “dad”, “”, “”] * 中找到"ball"，返回下标4. *  * 例子：在字符串数组 [“at”, “”, “”, “”, “”, “ball”, “car”, “”, “”, “dad”, “”, “”] * 中找到"ballcar"，查找失败，返回-1. *  */public class S11_5 {public static void main(String[] args) {String[] stringList = { "apple", "", "", "banana", "", "", "","carrot", "duck", "", "", "eel", "", "flower" };System.out.println(search(stringList, "apple"));}public static int search(String[] strings, String str) {if (strings == null || str == null || str.isEmpty()) {return -1;}return searchR(strings, str, 0, strings.length - 1);}public static int searchR(String[] strings, String str, int first, int last) {if(first > last){return -1;}int mid = (first+last)/2;// 找到离中心最近的不是空字符串的位置if(strings[mid].length() == 0 && mid<=last){int left = mid-1;int right = mid+1;while(true){if(left<first && right>last){return -1;}else if(right<=last && strings[right].length()!=0){mid = right;break;}else if(left>=first && strings[left].length()!=0){mid = left;break;}right++;left--;}}if(str.equals(strings[mid])){return mid;}else if(strings[mid].compareTo(str) > 0){return searchR(strings, str, first, mid-1);}else {return searchR(strings, str, mid+1, last);}}}