考虑这个数组:[["B","C","C","C","C","B","B","C","A","A"],["B","A","C","B","B","A","B","B","A","A"],["B","C","B","C","A","A","A","B","C","B"],["B","B","B","A","C","B","A","C","B","A"],["A","A","A","C","A","C","C","B","A","C"],["A","B","B","A","A","C","B","C","C","C"],["C","B","A","A","C","B","B","C","A"
我有一个像这样的树结构数据:[{id:54,name:123,children:[{id:54,name:123,children:[{id:154,name:1234,children[]...}]}]},{...}]我使用的是Angular2。据我所知,每当输入发生变化并且您的变化检测策略是onPush时,变化检测就会启动。.为了优化树结构更新(例如,在嵌套级别切换节点或更改此类节点的任何属性),我使用了Immutable.Immutable如何帮助我优化更新?我读到Immutable会在数据发生变化时重用旧数据的引用来构造新对象。如何有效地使用不可变数据结构来更新嵌套级别的节点?
在SO上和“网络”周围,它几乎已经完成了。但是我想知道是否有一种方法可以利用以下标准的最小/最大函数:Array.max=function(array){returnMath.max.apply(Math,array);};Array.min=function(array){returnMath.min.apply(Math,array);};所以我可以搜索对象数组,例如:functionVector(x,y,z){this.x=x;this.y=y;this.z=z;}varArrayVector=[/*lotsofdata*/];varmin_x=ArrayVector.x.min
我正在做一个我无法解决的练习。我需要通过买卖比特币来获得最大的累计利润。我有一个函数(A,Y),它在一段时间内接收一个A=不同价格的数组和一个Y=费用限制:注意:如果比特币以0的价格买入并以1的价格卖出,我们将损失A[1]-A[0]=7050-7200-Y=-200。因此,没有进行该运动。注意2:您当时只能拥有1个比特币。要卖,你必须先买。要购买,您需要之前一无所有或已售出。注3:运动需要是时间的结果。您不能在A[5]买入并在A[4]卖出注4:如果不能盈利,则返回0复杂度为O(N)A=[7200,7050,7300,7500,7440,7200,7300,7280,7400]//exp