这道题是否会让你想起快速排序中的分裂操作？只不过快速排序分开的标准是分大小分别存放而已。

//左偶右奇
void apartEvenOdd(ElemType Buff[],UINT32 len)
{
    ElemType temp;
    int head=0,tail = len-1;

    while (head<tail)
    {
        //寻找奇数
        if(Buff[head]%2==0)
        {
            head++;
        }

        //寻找偶数
        if(Buff[head]%2==1)
        {
            tail--;
        }

        if (head<tail)
        {
            //交换两个的顺序
            temp = Buff[head];
            Buff[head] = Buff[tail];
            Buff[tail] = temp;
        }
    }
}

python实现：

@print_result_after
def partition_even_odd(data):
    start = 0
    last = len(data) - 1

    while start <= last and is_even(data[start]):
        start += 1

    pivot = data[start]

    while start < last:
        while start < last:
            if is_even(data[last]):
                data[start] = data[last]
                start += 1
                break
            else:
                last -= 1

        while start < last:
            if not is_even(data[start]):
                data[last] = data[start]
                last -= 1
                break
            else:
                start += 1

    data[start] = pivot
    return data

def main():
    data = [3,0,1,2,4,5,6]
    partition_even_odd(data)