Question

我有以下代码。

function findMatch(array_1_small, array2_large) {
var ary = new Array();
for(i = 0;i < array2_large.length; i++)
{
  for(z = 0; z < array_1_small.length; z++)
  {
    if(array2_large[i] == array_1_small[z])
    {
     var idx = array2_large.indexOf(array2_large[i]);
      ary.push(idx);
    }
  }

}
return ary;
}

这需要以下数组。

var all_SMS_TO = ["0861932936", "0871355066", "0874132026", "0872908445", "0874132026"];
var all_FORM_NUMBERS = ["", "", "", "", "", "", "0871355066",""];

唯一的目的是找到匹配并在'all_form_numbers数组上返回匹配的索引。

在调用代码时

var a = findMatch(all_SMS_TO, all_FORM_NUMBERS);
console.log("Match Found " + a);

我得到以下输出。

Match Found: 6

这是正确的，但是当我将all_form_Numbers数组改为

时

var all_FORM_NUMBERS = ["", "0871355066", "", "", "", "", "0871355066",""];

我得到以下输出。

Match Found: 1,1

有人可以帮助我，所以它应该输出;

Match Found 1, 6.

感谢。

Answer 1

试试这个：

function findMatch(array_1_small, array2_large) {
var ary = new Array();
for(i = 0;i < array2_large.length; i++)
{
  for(z = 0; z < array_1_small.length; z++)
  {
    if(array2_large[i] == array_1_small[z])
    {
      ary.push(i);
    }
  }

}
return ary;
}

Answer 2

当你这样做时

var idx = array2_large.indexOf(array2_large[i]);

您正在数组array2_large中搜索值0871355066的索引两次，并且根据indexOf的定义，它将返回指定值的第一次出现的位置。

这就是为什么自第一次出现的索引以来索引值1两次。

对于解决方案只需将变量i值推入数组ary即可。这已经是循环中array2_large的索引值。

＆＃13;
＆＃13;
function findMatch(array_1_small, array2_large) { var ary = new Array(); for(i = 0;i < array2_large.length; i++) { for(z = 0; z < array_1_small.length; z++) { if(array2_large[i] == array_1_small[z]) { ary.push(i); } } } return ary; } var all_SMS_TO = ["0861932936", "0871355066", "0874132026", "0872908445", "0874132026"]; //var all_FORM_NUMBERS = ["", "", "", "", "", "", "0871355066",""]; var all_FORM_NUMBERS = ["", "0871355066", "", "", "", "", "0871355066",""]; var a = findMatch(all_SMS_TO, all_FORM_NUMBERS); console.log("Match Found " + a);
＆＃13;
＆＃13;
＆＃13;

Answer 3

你只需要推送索引i这里是固定代码（你也可以将数组声明为var res = [];

function findMatch(arraySmall, arrayLarge) {
  var res = []
  for (var i = 0; i < arrayLarge.length; i++) {
    for (var j = 0; j < arraySmall.length; j++) {
      if (arrayLarge[i] === arraySmall[j]) {
        res.push(i);
      }
    }
  }
  return res;
}

Answer 4

通过首先创建位置查找表，可以在O（n + m）运行时复杂度上解决这个问题。然后将每个元素从第一个数组映射到所有位置，并在Set中收集这些索引，只留下唯一值。

试试这个：

＆＃13;

var all_SMS_TO = ["0861932936", "0871355066", "0874132026", "0872908445", "0874132026"];
var all_FORM_NUMBERS = ["", "0871355066", "", "", "", "", "0871355066",""];

function findMatch(array_1_small, array2_large) {
    var positions = Array.from(array2_large.entries()).reduce((acc, t) => {
        var index = t[0]
        var element = t[1]
        if (!acc.hasOwnProperty(element)) {
            acc[element] = []
        }
        acc[element].push(index)
        return acc
    }, {})
    var result = new Set()
    array_1_small.forEach(x => {
        if (positions[x] === undefined) {
            return
        }
        positions[x].forEach(index => result.add(index))
    })
    return Array.from(result)
}

console.log("Match found: " + findMatch(all_SMS_TO, all_FORM_NUMBERS))

＆＃13;

JavaScript：比较两个数组并返回匹配的索引

4 个答案: