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]++;