如何删除子列表

Question

我怎样才能从列表中删除子表的所有事件，如

List(1, 2, 3, 4, 5, 6, 7, 4, 8, 9, 10, 5).removeSubList(4, 5) 

应该删除所有出现的（4,5）（顺序），所以它返回

List(1, 2, 3, 6, 7, 4, 8, 9, 10, 5) 

Answer 1

使用indexOfSlice递归溶液：

def removeSubList(l: List[Int], sublist: List[Int]): List[Int] = l.indexOfSlice(sublist) match { 
    case -1 => l 
    case index => removeSubList(l.patch(index, Nil, sublist.length), sublist) 
} 

// all of these print List(1 ,2 ,3): 
println(removeSubList(List(1,2,3), List(4,5))) 
println(removeSubList(List(1,2,3,4,5), List(4,5))) 
println(removeSubList(List(4,5,1,2,3), List(4,5))) 
println(removeSubList(List(4,5,1,2,4,5,3), List(4,5))) 

EDITED：

• （感谢@ corvus_192）恢复到使用indexOfSlice版本，而不是使用diff，而忽略子列表顺序。
• （感谢@The原型保罗）使用patch对清洁去除子表的

Answer 2

使用Tzach琐想法不同的实现：

def removeSubList[T](list: List[T], sublist: List[T]): List[T] = 
    if (list.containsSlice(sublist)) 
     removeSubList(list.diff(sublist), sublist) 
    else list 

Answer 3

上Tzach琐想法变但假定的OP想要的结果不包括曾经给定的子表。处理的情况下sublist是空

/** 
    * Return `s` with no occurrence of `target` present 
    * 
    * @param s 
    * @param target 
    * @return The sequence free of any occurrence of the target slice 
    */ 
    def removeCompletely[E](s: Seq[E], target: Seq[E]): Seq[E] = { 
    if (s.isEmpty || target.isEmpty) 
     s 
    else 
     removeCompletely0(s, target) 
    } 

    private def removeCompletely0[E](s: Seq[E], target: Seq[E]): Seq[E] = { 
    val sliceIdx = s.indexOfSlice(target) 
    if (sliceIdx >= 0) { 
     val patched = s.patch(sliceIdx, s take 0, target.size) 
     removeCompletely0(patched, target) 
    } 
    else { 
     s 
    } 
    }