# An array is defined to be odd-heavy if it contains at least one odd element

## 美国面试题

3. An array is defined to be odd-heavy if it contains at least one odd element and every element whose value is odd is greater than every even-valued element. So {11, 4, 9, 2, 8} is odd-heavy because the two odd elements (11 and 9) are greater than all the even elements. And {11, 4, 9, 2, 3, 10} is not odd-heavy because the even element 10 is greater than the odd element 9.
Write a function called isOddHeavy that accepts an integer array and returns 1 if the array is odd-heavy; otherwise it returns 0.
If you are programming in Java or C#, the function signature is int isOddHeavy(int[ ] a)
If you are programming in C or C++, the function signature is int isOddHeavy(int a[ ], int len) where len is the number of elements in the array

## java实现代码：

```package com.zzy;

/**
* 更多请关注: http://huamaodashu.com
* Created by huamaodashu on 20/07/2018.
*/
public class OddHeavy {
public static void main(String[] args) {

//        int[]a={2};
//        int[]a={1, 1, 1, 1, 1, 1};
//        int[]a={2, 4, 6, 8, 11};
int[]a={-2, -4, -6, -8, -11};
System.out.println(isOddHeavy(a));

}

public static int isOddHeavy(int[]a){

if(a.length==0){
return 0;
}else if(a.length==1){
if (isEven(a)==0){
return 1;
}else {
return 0;
}
}else {
int maxEvent=a;
int maxOdd=a;
for (int i = 0; i < a.length; i++) {
if(isEven(a[i])==1){

if(maxEvent<a[i]){
maxEvent=a[i];
}

}else if(isEven(a[i])==0){
if(maxOdd<a[i]){
maxOdd=a[i];
}

}else {
return 0;
}
}

if(maxEvent<maxOdd){
return 1;
}else if(maxEvent==maxOdd &&isEven(maxOdd)==0){
return 1;
}

}
return 0;
}

public static int isEven(int a){
if(a%2==0){
return 1;
}else {
return 0;
}
}

}
```

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