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数组元素判断算法挑战

方法一:

function every(collection, pre) {
  // Is everyone being true?
  var res=collection.every(function(item){
    return item[pre];
  });
  return res;
}


every([{"user": "Tinky-Winky", "sex": "male"}, {"user": "Dipsy", "sex": "male"}, {"user": "Laa-Laa", "sex": "female"}, {"user": "Po", "sex": "female"}], "sex");

方法二:

function every(collection, pre) {
  var isTruthy = true;
  // Is everyone being true?
  collection.forEach(function(obj){
    if(!obj[pre]||obj[pre]===''){
      isTruthy = false;
      return false;
    }
  });
  return isTruthy;
}


every([{"user": "Tinky-Winky", "sex": "male"}, {"user": "Dipsy", "sex": "male"}, {"user": "Laa-Laa", "sex": "female"}, {"user": "Po", "sex": "female"}], "sex");
二进制转字符算法挑战
函数迭代可选参数算法挑战
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