# Define an array to be packed if all its values are positive, each value n appears n times and all equal values are in consecutive locations.

Define an array to be packed if all its values are positive, each value n appears n times and all equal values are in consecutive locations. So for example, {2, 2, 3, 3, 3} is packed because 2 appears twice and 3 appears three times. But {2, 3, 2, 3, 3} is not packed because the 2s are not in consecutive locations. And {2, 2, 2, 3, 3, 3} is not packed because 2 appears three times.
Write a method named isPacked that returns 1 if its array argument is packed, otherwise it returns 0. You may assume that the array is not null
If you are programming in Java or C#, the function signature is int isPacked(int[ ] a)
If you are programming in C++ or C, the function signature is int isPacked(int a[ ], int len) where len is the length of the array.

java代码实现：

```package com.zzy;

/**
* 更多请关注: http://huamaodashu .com
* Created by huamaodashu on 20/07/2018.
*/
public class Packed {

public static void main(String[] args) {

//        int [] a={2, 2, 1};
//        int [] a={2, 2, 1, 2, 2};
//        int [] a={4, 4, 4, 4, 1,2, 2, 3, 3, 3};
//        int [] a={7, 7, 7, 7, 7,  7, 7, 1};
//        int [] a={7, 7, 7, 7, 1,  7, 7, 7};
//        int [] a={7, 7, 7, 7, 7,  7, 7};
int [] a={};

System.out.println(isPacked(a));

}

public static int isPacked(int[] a){
if(a.length>0){
int tmp=a[0];
int count=1;

for (int i = 1; i < a.length; i++) {
if(tmp==a[i]){
count++;
int countgroup=0;
for (int j = 0; j < a.length; j++) {
if(a[j]==tmp){
countgroup++;
}
}
if(tmp!=countgroup){
return 0;
}
}else { //不等于了,

if(tmp!=count){
System.out.println("ai="+i+"count="+count);
return 0;
}
tmp=a[i];
count=1;

}
}

return 1;
}
return 1;
}
}
```