题目1 : 分隔相同整数

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

描述

给定一个包含N个整数的数组A。你的任务是将A重新排列，使得任意两个相等的整数在数组中都不相邻。  

如果存在多个重排后的数组满足条件，输出字典序最小的数组。  

这里字典序最小指：首先尽量使第一个整数最小，其次使第二个整数最小，以此类推。

输入

第一行包含一个整数N，表示数组的长度。(1 <= N <= 100000)  

第二行包含N个整数，依次是 A1, A2, ... AN。(1 <= Ai <= 1000000000)

输出

输出字典序最小的重排数组。如果这样的数组不存在，输出-1。

样例输入

4  
2 1 3 3

样例输出

1 3 2 3

import java.util.*;
import java.io.*;
public class Main{
	public static void main(String[] args)throws Exception {
		Scanner sc = new Scanner(System.in);
		while(sc.hasNext()){
			TreeMap<Integer,Integer> numbers = new TreeMap<Integer,Integer>();
			int n = sc.nextInt();
			int temp;
			for(int i = 0; i < n; ++i){
				temp = sc.nextInt();
				if(numbers.containsKey(temp)){
					numbers.replace(temp,numbers.get(temp)+1);
				}
				else{
					numbers.put(temp,1);
				}
			}
			
			boolean cantFound = false;
			for(Map.Entry<Integer,Integer> pair: numbers.entrySet()){
				if(pair.getValue() > (n+1)/2){
					System.out.println(-1);
					cantFound = true;
					break;
				}
			}
			if(cantFound){
				continue;
			}
			
			int last = -1;
			int key;
			Integer higherKey,higherValue;
			ArrayList<Integer> result = new ArrayList<Integer>(n);
			while(numbers.size() > 0){
				key = numbers.firstKey();
				if(key != last){
					addKey(key,result,numbers);
					last = key;
				}
				else{
					higherKey = numbers.higherKey(key);
					if(higherKey == null){
						
						findPlace(key,result,numbers);
					}
					else{
						addKey(higherKey,result,numbers);
						last = higherKey;
					}
				}
			}
	
			System.out.printf("%d",result.get(0));
			for(int i = 1; i < result.size(); ++i){
				System.out.printf(" %d",result.get(i));
			}
			System.out.printf("\n");
		}
	}
	public static void addKey(int key, ArrayList<Integer> result, TreeMap<Integer,Integer> numbers){
		result.add(key);
		int value = numbers.get(key);
		if(value > 1){
			numbers.replace(key,value-1);
		}
		else{
			numbers.remove(key);
		}
	}
	public static void findPlace(int key, ArrayList<Integer> result, TreeMap<Integer,Integer> numbers){
		//the elements left in numbers are the same number,so insert them one by one.
		int insertPos = result.size()-2;//size will change after insert operation,be careful.
		int amountOfKey = numbers.get(key);
		int replacePos = result.size()-2;
		for(int i = 0; i < amountOfKey; ++i){
			while(insertPos > 0){
				if(result.get(insertPos) != key && result.get(insertPos-1) != key){
					if(result.get(insertPos) > result.get(insertPos-1)){
						for(int j = result.size()-2; j > insertPos; --j){
							if(result.get(j) == result.get(insertPos-1)){
								int temp = result.get(j);
								result.set(j,result.get(insertPos));
								result.set(insertPos,temp);
								break;
							}
						}
						
					}
					numbers.replace(key,numbers.get(key)-1);
					result.add(insertPos,key);
					insertPos = insertPos - 1; 
					break;
				}
				--insertPos;
			}
		}
		//the last one(if exist) to be insert in first place.
		//only if numbers.get(k) == (n+1)/2
		if(numbers.get(key) > 0){
			result.add(0,key);
		}
		numbers.remove(key);

	
	}
}