Question

我需要在一组数组中找到第一个公共元素。阵列的数量可以变化，但它们总是按顺序排列（小 - >大）。我的数组都是myObj的属性。

这是我到目前为止所做的：

function compare(myObj,v,j) {
  if (myObj[j].indexOf(v)>-1) return true;
  else return false;
}

function leastCommon ([1,5]) {
  var myObj = { //This is filled by code, but the finished result looks like this
    1: [1, 2,...,60,...10k]
    2: [2, 4,...,60,...20k]
    3: [3, 6,...,60,...30k]
    4: [4, 8,...,60,...40k]
    5: [5,10,...,60,...50k]
  };
  var key = [1,2,3,4,5]; //also filled by code
  var lcm = myObj[key[key.length-1]].forEach(function(v) { //Iterate through last/largest multiple array
  var j=key[key.length-2];
  while (j>=0) {
    if (compare(myObj,v,j)) {  //check to see if it is in the next lower array, if yes, check the next one.
      j--;
    }
    if (j>0 && (compare(myObj,v,j+1))===true) return v;  //before the loop exits, return the match
   }  
 });
return lcm;
}

我不确定是什么问题，但它返回未定义。

注意：是的，我知道forEach返回undefined，我尝试修改我的代码，并且我的编辑器出现“潜在的无限循环”错误。修改后的代码如下所示：

function leastCommon ([1,5]) {
  var myObj = { //This is filled by code, but the finished result looks like this
    1: [1, 2,...,60,...10k]
    2: [2, 4,...,60,...20k]
    3: [3, 6,...,60,...30k]
    4: [4, 8,...,60,...40k]
    5: [5,10,...,60,...50k]
  };
  var key = [1,2,3,4,5]; //also filled by code
  var lcm = 0;
  myObj[key[key.length-1]].forEach(function(v) { //Iterate through last/largest multiple array
    var j=key[key.length-2];
    while (j>=0) {
      if (compare(myObj,v,j)) {  //check to see if it is in the next lower array, if yes, check the next one.
        j--;
      }
      if (j>0 && (compare(myObj,v,j+1))===true) lcm = v;  //before the loop exits, set lcm = v
    }  
  });
return lcm;
}

Answer 1

我不会使用forEach，因为当您找到第一个匹配/失败时无法退出该方法。你需要保持循环。相反，你应该查看每个和indexOf的常规for循环。此代码还假定数组已排序，因此最小的数字首先出现。如果没有，带有数组克隆的简单sort（）可以解决这个问题。

//pass in arrays, assumes array is sorted
function getFirstCommon (arrs) {
  //Loop over the elements in the first array
  for (var i=0; i<arrs[0].length; i++) {
    //get the value for the current index
    var val = arrs[0][i];
    //make sure every array has the value
    //if every does not find it, it returns false
    var test = arrs.every( function (arr) {
      //check the array to see if it has the element
      return arr.indexOf(val)!==-1;  
    });
    //If we find it, than return the current value
    if (test) {
      return val;
    }   
  }
  //if nothing was found, return null
  return null; 
}

//test numbers
var nums = [
   [1,2,3,4,5,6],
   [2,3,4,5,6],
   [3,4,5,6],
   [4,5,6,7,8,9],
   [6,7,8,9]
];
  
console.log(getFirstCommon(nums));  //6
  
var nums2 = [
   [1,2,3,4,5,6],
   [2,3,4,5,6],
   [3,4,5,6],
   [4,5,6,7,8,9],
   [5,7,8,9]
];
  
console.log(getFirstCommon(nums2));  //5

var nums3 = [
   [1,2,3,4,5,6],
   [7,8,9,10,11,12],
   [7,8,9,10,11,12]
];
  
console.log(getFirstCommon(nums3));  //null

可以在不自行检查的情况下改进代码

Answer 2

首先，你确实有一个无限循环。如果第一个compare()失败，您永远不会减少j，并继续检查相同的数字。

第二：你key数组永远不会减少，所以你总是比较最后两个数组。

Javascript中跨多个数组的第一个常见元素

2 个答案: