G面经prepare: Reorder String to make duplicates not consecutive

字符串重新排列，让里面不能有相同字母在一起。比如aaabbb非法的，要让它变成ababab。给一种即可

Greedy:

跟FB面经Prepare task Schedule II很像，记录每个char出现次数，然后用最大堆，把剩下char里面出现次数多的优先Poll出来组建新的string

如果poll出来的char跟上一个相同，则用一个queue暂时存一下

我觉得时间复杂度：O(N) + O(KlogK) + O(NlogK) = O(NlogK) ,where K is the number of different character in the string

 1 package ReorderString;
 2 import java.util.*;
 3 
 4 public class Solution {
 5     class Element {
 6         char val;
 7         int appear;
 8         public Element(char value) {
 9             this.val = value;
10             this.appear = 1;
11         }
12     }
13     
14     public String reorder(String str) {
15         Element[] summary = new Element[26];
16         for (int i=0; i<str.length(); i++) {
17             char cur = str.charAt(i);
18             if (summary[(int)(cur-'a')] == null) {
19                 summary[(int)(cur-'a')] = new Element(cur);
20             }
21             else {
22                 summary[(int)(cur-'a')].appear++;
23             }
24         }
25         PriorityQueue<Element> queue = new PriorityQueue<Element>(11, new Comparator<Element>() {
26             public int compare(Element e1, Element e2) {
27                 return e2.appear - e1.appear;
28             }
29         });
30         
31         for (Element each : summary) {
32             if (each != null) {
33                 queue.offer(each);
34             }
35         }
36         Queue<Element> store = new LinkedList<Element>();
37         StringBuffer res = new StringBuffer();
38         while (!queue.isEmpty() || !store.isEmpty()) {
39             if (!queue.isEmpty()) {
40                 Element cur = queue.poll();
41                 if (res.length()==0 || cur.val!=res.charAt(res.length()-1)) {
42                     res.append(cur.val);
43                     cur.appear--;
44                     if (cur.appear > 0) store.offer(cur);
45                     while (!store.isEmpty()) {
46                         queue.offer(store.poll());
47                     }
48                 }
49                 else { //cur.val equals last char in res
50                     store.offer(cur);
51                 }
52             }
53             else { //store is not empty but queue is empty
54                 res = new StringBuffer();
55                 res.append(-1);
56                 return res.toString();
57             }
58         }
59         return res.toString();
60     }
61     
62 
63     /**
64      * @param args
65      */
66     public static void main(String[] args) {
67         // TODO Auto-generated method stub
68         Solution sol = new Solution();
69         String res = sol.reorder("aaabbba");
70         System.out.println(res);
71     }
72 
73 }