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

 package ReorderString;
 import java.util.*;

 public class Solution {
     class Element {
         char val;
         int appear;
         public Element(char value) {
             this.val = value;
             this.appear = 1;
         }
     }

     public String reorder(String str) {
         Element[] summary = new Element[26];
         for (int i=0; i<str.length(); i++) {
             char cur = str.charAt(i);
             if (summary[(int)(cur-'a')] == null) {
                 summary[(int)(cur-'a')] = new Element(cur);
             }
             else {
                 summary[(int)(cur-'a')].appear++;
             }
         }
         PriorityQueue<Element> queue = new PriorityQueue<Element>(11, new Comparator<Element>() {
             public int compare(Element e1, Element e2) {
                 return e2.appear - e1.appear;
             }
         });

         for (Element each : summary) {
             if (each != null) {
                 queue.offer(each);
             }
         }
         Queue<Element> store = new LinkedList<Element>();
         StringBuffer res = new StringBuffer();
         while (!queue.isEmpty() || !store.isEmpty()) {
             if (!queue.isEmpty()) {
                 Element cur = queue.poll();
                 if (res.length()==0 || cur.val!=res.charAt(res.length()-1)) {
                     res.append(cur.val);
                     cur.appear--;
                     if (cur.appear > 0) store.offer(cur);
                     while (!store.isEmpty()) {
                         queue.offer(store.poll());
                     }
                 }
                 else { //cur.val equals last char in res
                     store.offer(cur);
                 }
             }
             else { //store is not empty but queue is empty
                 res = new StringBuffer();
                 res.append(-1);
                 return res.toString();
             }
         }
         return res.toString();
     }

     /**
      * @param args
      */
     public static void main(String[] args) {
         // TODO Auto-generated method stub
         Solution sol = new Solution();
         String res = sol.reorder("aaabbba");
         System.out.println(res);
     }

 }