[7kyu]Alan Partridge I - Partrid

该算法题来自于 codewars【语言: javascript】，翻译如有误差，敬请谅解~

• 任务

• 给出一系列术语组成的数组，如果其中任何一个术语与 Alan Partridge 有关，就返回 Mine's a Pint！

• Mine's a Pint之后！的数量，由所提供的数组（x）与的Alan Partridge相关的术语个数确定。相关术语为：
  Partridge
  PearTree
  Chat
  Dan
  Toblerone
  Lynn
  AlphaPapa
  Nomad
• 如果没有任何一个术语与Alan Partridge相关，请返回Lynn, I've pierced my foot on a spike!!
• 例如：
  part(['Grouse', 'Partridge', 'Pheasant']) // 'Mine's a Pint!'
  part(['Pheasant', 'Goose', 'Starling', 'Robin']) // 'Lynn, I've pierced my foot on a spike!!'
  part(['Grouse', 'Partridge', 'Pheasant', 'Goose', 'Starling', 'Robin', 'Thrush', 'Emu', 'PearTree', 'Chat', 'Dan', 'Square', 'Toblerone', 'Lynn', 'AlphaPapa', 'BMW', 'Graham', 'Tool', 'Nomad', 'Finger', 'Hamster']) // 'Mine's a Pint!!!!!!!!'

• 解答
• 其一

function part(x){
      let len = x.join('').replace(/Partridge|PearTree|Chat|Dan|Toblerone|Lynn|AlphaPapa|Nomad/g,'#').split('#').length - 1;
      let str = "Lynn, I've pierced my foot on a spike!!";
      if (len) str = "Mine\'s a Pint" + '!'.repeat(len);
      return str;
}

• 其二

function part(x){
      var count = x.filter(e => ['Partridge','PearTree','Chat','Dan','Toblerone','Lynn','AlphaPapa','Nomad'].indexOf(e) != -1).length;
      return count > 0 ? 'Mine\'s a Pint' + '!'.repeat(count) : 'Lynn, I\'ve pierced my foot on a spike!!';
}