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

 1 package SortStringBasedOnAnother;
 2 import java.util.*;
 3 
 4 public class Solution {
 5     public class Sortable implements Comparable<Sortable> {
 6     
 7         private Character content;
 8         private int order;
 9 
10         public Sortable(char str, String sortingOrder) {
11             this.content = str;
12             this.order = sortingOrder.indexOf(str);
13             if (this.order == -1) this.order = Integer.MAX_VALUE;
14         }
15 
16         public int compareTo(Sortable another) {
17             if (this.order > another.order) {
18                 return 1;
19             } else if (this.order == another.order) {
20                 return 0;
21             } else {
22                 return -1;
23             }
24         }
25         
26         public String toString() {
27             return String.valueOf(content);
28         }
29 
30     }
31     
32     public void sort(String order, String str) {
33         Sortable[] array = new Sortable[str.length()];
34         for (int i = 0; i < str.length(); i++) {
35             array[i] = new Sortable(str.charAt(i), order);
36         }
37         Collections.sort(Arrays.asList(array)); //Arrays.sort(array);
38         for (int i = 0; i < array.length; i++) {
39             System.out.print(array[i]);
40         }
41         
42     }
43 
44     public static void main(String[] args) {
45             Solution sol = new Solution();
46             String order = "dfbcae";
47             String str = "abcdkkeeabc";
48             sol.sort(order, str);
49     }
50 }

 

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

 1 package SortStringBasedOnAnother;
 2 import java.util.*;
 3 
 4 public class Solution2 {
 5     public void sort(String order, String str) {
 6         int[] count = new int[order.length()];
 7         Queue<Character> notInOrder = new LinkedList<Character>();
 8         StringBuffer sb = new StringBuffer();
 9         for (int i=0; i<str.length(); i++) {
10             boolean matched = false;
11             char cur = str.charAt(i);
12             for (int j=0; j<order.length(); j++) {
13                 if (cur == order.charAt(j)) {
14                     count[j]++;
15                     matched = true;
16                     break;
17                 }
18             }
19             if (!matched) notInOrder.offer(cur);
20         }
21         for (int i=0; i<count.length; i++) {
22             while (count[i] > 0) {
23                 sb.append(order.charAt(i));
24                 count[i]--;
25             }
26         }
27         while (!notInOrder.isEmpty()) {
28             sb.append(notInOrder.poll());
29         }
30         System.out.println(sb.toString());
31     }
32     
33 
34     /**
35      * @param args
36      */
37     public static void main(String[] args) {
38         // TODO Auto-generated method stub
39         Solution2 sol = new Solution2();
40         String order = "dfbcae";
41         String str = "akbcdeeabc";
42         sol.sort(order, str);
43 
44     }
45 
46 }

12-19行可以用IndexOf替代

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