Question

我试图弄清楚如何使用JavaScript ex计算数组中2个重复数字之间的距离：

    ["2","3","4","2"] => 3 (2 is a duplicate number)
    ["1","4","5","5"] => 1 (5 is a duplicate number)
    ["1","4","5","9","4"] => 3 (4 is a duplicate number)

Answer 1

假设您的数组中任何给定值的出现次数不会超过两次，我们可以采取以下步骤： -

遍历数组项并获取每个项目的所有索引
将项目添加到encountered数组中，这有助于您确定是否已搜索该项目，
如果该项目有2次出现，请获取索引之间的差异，这将是您的步数。

在Javascript中，以下代码将为您执行相同的操作。

＆＃13;

var temp = [2,3,4,2,5,4,1,3];
var encountered = [];

//This function gets you all the indexes of `val` inside `arr` array.
function getAllIndexes(arr, val) {
  var indexes = [], i;
  for(i = 0; i < arr.length; i++)
    if (arr[i] === val)
      indexes.push(i);
  return indexes;
}

for(var i=0;i<temp.length;i++) {
    if(encountered[temp[i]]) continue; 
    else {
        var indexes = getAllIndexes(temp, temp[i]);
        encountered[temp[i]] = true;
        if(indexes.length>1) {
            var steps = indexes[1]-indexes[0];
            $('.container').append('Duplicate item: '+temp[i]+' steps: '+ steps+'<br/>');
        }
    }
}

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.min.js"></script>
<div class="container"></div>

＆＃13;

在这里工作JSFiddle - ＆gt; https://jsfiddle.net/dmnuwas5/

Answer 2

希望这个伪代码能让你走上正确的道路：

values = original array
distances = {}
firstIndex = {}

loop over values with index i {
  item = values[i]
  if firstIndex[item] >= 0 { // this is important because the first index could be 0
    if item does not exist in distances object { // This would be different if you are going for max distance instead of min distance in the case of multiple of the same item
      distances[item] = i - firstIndex[item]
    }
  } else {
    firstIndex[item] = i
  }
}

Answer 3

这是多次出现同一项目并计算至少最小距离的建议。

＆＃13;

function getDistance(array) {
    var o = {}, r = {};
    array.forEach(function (a, i) {
        o[a] = o[a] || [];
        o[a].push(i);
    });
    Object.keys(o).forEach(function (k) {
        if (o[k].length > 1) {
            r[k] = Math.min.apply(null, o[k].reduce(function (r, a, i, aa) {
                i && r.push(a - aa[i - 1]);
                return r;
            }, []));
        }
    });
    return r;
}

document.write('<pre>' + JSON.stringify(getDistance([2,3,4,2,5,4,1,3]), 0, 4) + '</pre>');
document.write('<pre>' + JSON.stringify(getDistance(["2", "3", "4", "2", "3", "5", "6", "2"]), 0, 4) + '</pre>');

＆＃13;

Answer 4

满足您要求的最小示例。它返回一个对象，其中重复的数字作为键，距离为值。

function sift(arr) {
  for (var i =  0, l = arr.length, out = {}; i < l; i++) {
    var key = arr.shift();
    var index = arr.indexOf(key);
    if (index > -1 && !out[key]) out[key] = index + 1;
  }
  return out;
}

sift(["2","3","4","2"]); // { 2: 3 }
sift(["1","4","5","5"]); // { 5: 1 }
sift(["1","4","5","9","4"]); // { 4: 3 }
sift(["2","3","4","2","3","5","6","2"]); // { 2: 3, 3: 3 }
sift(["6","2","3","4","2","3","5","6","2"]); // { 2: 3, 3: 3, 6: 7 }

DEMO

Answer 5

用于查找重复元素之间距离的Java代码。如果存在多个重复元素，则将返回重复元素对之间的最小距离。此外，假设重复元素出现两次（不超过两次）
时间复杂度：O（n）
空间复杂度：O（n）

public int minDuplicateDistance(int[] arr) {
    Map<Integer, Integer> dups = new HashMap<Integer, Integer>();
    int minDistance = Integer.MAX_VALUE;
    for (int x = 0; x < arr.length; x++) {
        if (dups.containsKey(arr[x])) {
            int index = dups.get(arr[x]);
            int distance = x - index;
            minDistance = minDistance > distance ? distance : minDistance;
        } else {
            dups.put(arr[x], x);
        }
    }
    return minDistance;
}

数组中2个重复数字之间的距离

5 个答案: