小编Evg*_*niy的帖子

例如,我们有系列1,2,3,4,5.我们采用每3个元素=> 3,1,5,2,4(选择的元素不应该保留,我们可以采用系列不为空).圈子双向链表的朴素实现不是表现的好主意.你能给我一个建议,哪些数据结构和算法更适用？

你有许多已知重量为w1,...,wn的宝石.编写一个程序,将石头重新排列成两堆,使两桩之间的重量差异最小.我有dp算法:

int max(int a, int b){
    return a >= b ? a : b;
}

int diff(int* weights, int number_elements, int capacity){
    int **ptrarray = new int* [capacity + 1]; 

    for (int count = 0; count <= capacity; count++) {
        ptrarray[count] = new int [number_elements + 1];
    }

    for (int i = 0; i <= number_elements; ++i){
        ptrarray[0][i] = 0;
    }

    for (int i = 0; i <= capacity; ++i){
        ptrarray[i][0] = 0;
    }

    for (int i = 1; i …

我正在使用Twitter API按照文档实现OAuth: 仅应用程序身份验证 我使用Consumer key&Consumer Secret创建了POST请求,如下所示:

POST http://api.twitter.com/oauth2/token
Authorization:  Basic T3k2TlA4SnpCVGRiNXlFTUt2dGswamJGSTpiSlpEYm1xeExyeDJKU25JbUplcWdlQTJkREcwZXg0bUNtOUdGTGJ1TGF3TkZkNkxqNg==
Content-Type: application/x-www-form-urlencoded;charset=UTF-8
grant_type=client_credentials

我得到了一个有以下错误的回复:

"{"errors":[{"code":99,"label":"authenticity_token_error","message":"无法验证您的凭据"}]}"

虽然,我从我的应用程序设置中获取数据

type Occurrences = List[(Char, Int)]

val lard = List(('a', 1), ('d', 1), ('l', 1), ('r', 1))
val r = List(('r', 1))


def subtract2(x: Occurrences, y: Occurrences): Occurrences =
  x.foldLeft(List[(Char, Int)]())({case (acc, list) if(!(y.contains(list))) => acc.::(list)})

subtract2(lard, r)

抛出

scala.MatchError: (List(),(a,1)) (of class scala.Tuple2)
at forcomp.A$A183$A$A183$$anonfun$subtract2$1.apply(temp.sc:11)
at forcomp.A$A183$A$A183$$anonfun$subtract2$1.apply(temp.sc:11)
at scala.collection.LinearSeqOptimized$class.foldLeft(temp.sc:120)
at scala.collection.immutable.List.foldLeft(temp.sc:80)
at forcomp.A$A183$A$A183.subtract2(temp.sc:11)
at #worksheet#.#worksheet#(temp.sc:14)

这个错误的原因是什么？我想,所有问题都有缺点,但更确切地说我不知道