Question

我正在尝试查找数组中元素的单次出现。但这仅显示1个元素。逻辑错在哪里？

function findSingle(array){
  var arrayCopy = array.slice(0);
  var x;
  var y = [];

  for (var i = 0; i < array.length; i++) {
    x = arrayCopy.splice(i, 1)
    if(arrayCopy.includes(array[i]) === false){
      console.log(array[i] + " is single")
    }
    arrayCopy = arrayCopy.concat(x)
   }
 }

findSingle([1, 3, 3, 6])

Answer 1

您可以使用双https://github.com/mabbashm110/Sprint-Challenge--Responsive-Less来删除出现多次的数字：

    function findSingle(arr) {
      return arr.filter(i => arr.filter(j => i === j).length === 1)
    }
    
    const result = findSingle([1, 3, 3, 6, 8, 4, 6])
    console.log(result) // [1, 8, 4]

请记住，由于Array.filter()，这仅适用于数字和其他primitives。

Answer 2

我添加了一些console.logs以查看发生了什么，问题是您正在更改arrayCopy中元素的顺序。因此，6永远不会被选中。

Checking for [ 1 ]
arrayCopy is [ 3, 3, 6 ]
1 is single
After adding to arrayCopy [ 3, 3, 6, 1 ]
Checking for [ 3 ]
arrayCopy is [ 3, 6, 1 ]
After adding to arrayCopy [ 3, 6, 1, 3 ]
Checking for [ 1 ]
arrayCopy is [ 3, 6, 3 ]
After adding to arrayCopy [ 3, 6, 3, 1 ]
Checking for [ 1 ]
arrayCopy is [ 3, 6, 3 ]
After adding to arrayCopy [ 3, 6, 3, 1 ]

您可能可以使用频率图来查找每个元素中的出现次数，然后过滤仅出现一次的键。

function findSingle(array){
  var freqs = {};
  array.forEach(n => {
    if (!(n in freqs)) freqs[n] = 1;
    else freqs[n] += 1;
  });
  return Object.keys(freqs).filter(k => freqs[k] === 1);
}

Answer 3

这可以通过创建一个可以映射元素及其出现的对象来完成。这是一个代码

function findSingle(arr){
    var counts = {};
    var singles = [];
    for (var i = 0; i < arr.length; i++) {
        var num = arr[i];
        counts[num] = counts[num] ? counts[num] + 1 : 1;
    }
    for(var num in counts) {
        if(counts[num] == 1)
            singles.push(num);
    }
    return singles.map(x => Number(x));
}

findSingle([1, 3, 3, 6])的输出将为

[1, 6]

注意，它也可以用于字符串，但可以是数字。例如，["1", "3", "3", "6"]

查找数组中元素的单次出现

3 个答案: