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);
    }

}