从两个数组中删除重复的数字

removing duplicate numbers out of both arrays

我知道您可以从数组中删除 1 个重复数字,但是如果数字重复,有没有办法删除该数字?到目前为止,下面的代码是我所拥有的。如果它们相等,我想使用 for 循环从数组中删除数字,但我认为我的编码不正确或不完整。我想要 return [3, 4, 5]

function sym(args) {
  var array = [];
  var join;
    for(var i = 0; i < arguments.length; i++){
      array.push(arguments[i]);
      join = array[0].concat(array[1]); 
    } 
  join.sort();
    for(var j = 0; j < join.length; j++) {
      if(join[j] === join[j+1]) {

        var removed = join.splice(j, join[j+2]);                 
        console.log(removed);
      }

    }  

  return join;

}

sym([1, 2, 3], [5, 2, 1, 4]);

这是我的看法

function sym() {
  var vals = {};
  var rarray= [];
  var a=arguments;
  for (var i=0,l=a.length;i<l;i++) {
    if (a[i] instanceof Array) {
      for (var n=0,ln=a[i].length;n<ln;n++) {
        vals[a[i][n]]=vals[a[i][n]]||[];
        vals[a[i][n]].push(a[i][n]);
      }
    }
  }
  for (var i in vals) {
    if (vals.hasOwnProperty(i)) {
      if (vals[i].length===1)
        rarray.push(i);
    }
  }
  return rarray;
}

示例:

sym([1, 2, 3], [5, 2, 1, 4]);
// return: ["3", "4", "5"]

sym([1, 2, 3], [5, 2, 1, 4],[4,6,7,8],[8,4]);
// ["3", "5", "6", "7"]

sym([1,2],[1]);
// return: ["2"]

我喜欢 Crayon Violent 的解决方案,但我不明白维护重复项数组的意义,因为您可以简单地对它们进行计数。

这大大提高了性能 (jsperf),同时还简化了代码。

function sym() {
  var occurrences = {};
  var inputArrays = arguments;
  var uniqueItems = [];

  function addOccurrences(arr) {
    for (var i = 0, len=arr.length; i < len; i++) {
      occurrences[arr[i]] = 1 + (occurrences[arr[i]] || 0);
    }
  }

  for (var i=0, len=inputArrays.length; i < len; i++) {
    if (inputArrays[i] instanceof Array) {
      addOccurrences(inputArrays[i]);
    }
  }

  for (var item in occurrences) {
    if (occurrences[item] === 1) {
        uniqueItems.push(item);
    }
  }
  return uniqueItems;
}

如果您的项目中恰好有下划线或lodash,可以做得更好:

function sym() {
  var inputArrays = _.filter(arguments, _.isArray);
  var occurrences = {};

  function addOccurrences(arr) {
    _.forEach(arr, function(item){
      occurrences[item] = 1 + (occurrences[item] || 0);
    });
  }

  _.forEach(inputArrays, addOccurrences);

  // Select the items with one occurence, return the keys (items)
  return _.filter(_.keys(occurrences), function(item){ 
    return occurrences[item] === 1; 
  });
}
var sym = function (ar1, ar2) {
    return ar1
        .concat(ar2)
        .sort(function (a, b) { return a - b; })
        .filter(function (elem, i, ar) {
            return ar[i-1] !== elem && ar[i+1] !== elem;
        });
    }