G面经prepare: Sort String Based On Another

Given a sorting order string, sort the input string based on the given sorting order string. Ex sorting order string -> dfbcae
Input string -> abcdeeabc
output -> dbbccaaee

法一：Comparable

sample Input:

String order = "dfbcae";
String str = "akbcdeeabc";

Sample Output: dbbccaaeekk

要注意13行，indexOf()没有matched到是return -1, 要把它设为最大

 package SortStringBasedOnAnother;
 import java.util.*;

 public class Solution {
     public class Sortable implements Comparable<Sortable> {

         private Character content;
         private int order;

         public Sortable(char str, String sortingOrder) {
             this.content = str;
             this.order = sortingOrder.indexOf(str);
             if (this.order == -1) this.order = Integer.MAX_VALUE;
         }

         public int compareTo(Sortable another) {
             if (this.order > another.order) {
                 return 1;
             } else if (this.order == another.order) {
                 return 0;
             } else {
                 return -1;
             }
         }

         public String toString() {
             return String.valueOf(content);
         }

     }

     public void sort(String order, String str) {
         Sortable[] array = new Sortable[str.length()];
         for (int i = 0; i < str.length(); i++) {
             array[i] = new Sortable(str.charAt(i), order);
         }
         Collections.sort(Arrays.asList(array)); //Arrays.sort(array);
         for (int i = 0; i < array.length; i++) {
             System.out.print(array[i]);
         }

     }

     public static void main(String[] args) {
             Solution sol = new Solution();
             String order = "dfbcae";
             String str = "abcdkkeeabc";
             sol.sort(order, str);
     }
 }

方法二：（Better）counting sort

If taking extra mem in allowed. Take an int array with size same as "sorting order string" that will maintain a count. now iterate the input string & keep on incrementing the corresponding char count in this new array.

记得要用一个queue记录没有在order里面的str的元素

 package SortStringBasedOnAnother;
 import java.util.*;

 public class Solution2 {
     public void sort(String order, String str) {
         int[] count = new int[order.length()];
         Queue<Character> notInOrder = new LinkedList<Character>();
         StringBuffer sb = new StringBuffer();
         for (int i=0; i<str.length(); i++) {
             boolean matched = false;
             char cur = str.charAt(i);
             for (int j=0; j<order.length(); j++) {
                 if (cur == order.charAt(j)) {
                     count[j]++;
                     matched = true;
                     break;
                 }
             }
             if (!matched) notInOrder.offer(cur);
         }
         for (int i=0; i<count.length; i++) {
             while (count[i] > 0) {
                 sb.append(order.charAt(i));
                 count[i]--;
             }
         }
         while (!notInOrder.isEmpty()) {
             sb.append(notInOrder.poll());
         }
         System.out.println(sb.toString());
     }

     /**
      * @param args
      */
     public static void main(String[] args) {
         // TODO Auto-generated method stub
         Solution2 sol = new Solution2();
         String order = "dfbcae";
         String str = "akbcdeeabc";
         sol.sort(order, str);

     }

 }

12-19行可以用IndexOf替代

             char cur = str.charAt(i);
             int index = order.indexOf(cur);
             if (index == -1) notInOrder.offer(cur);
             else count[index]++;