使用Javascript的数组中两个数字之间的最小距离

Question

我正在寻找一种更好的解决方案来获得数组中 2 个元素之间的最小距离。 输入：arr[] = {3, 5, 4, 2, 6, 5, 6, 6, 5, 4, 8, 3}, x = 3, y = 6 输出：3 和 6 之间的最小距离为 4。

我在 JS 中有这段代码，它现在可以正常工作。 我正在寻找更好的代码来实现相同的目标。 谢谢！！

<script>
var numbers= ["2", "3", "5","7","1","2","3","4","8"];
var x ="5";
var y ="8";
var firstIndex = numbers.indexOf(x);

var minD = numbers.length;

var x= numbers.forEach(function(item,index){

if((item == x) || (item == y))
{
    if((index != firstIndex) && (index-firstIndex < minD))
    {
    minD = index-firstIndex;
    firstIndex = index;
    }
    else
    {
    firstIndex = index;
    }
 }

});
alert(minD);

document.getElementById("demo").innerHTML = minD;
</script>

Answer 1

var xs=array.reduce((arr,el,i)=>(!(el===x)||arr.push(i),arr),[]);
var ys=array.reduce((arr,el,i)=>(!(el===y)||arr.push(i),arr),[]);
var lowest= xs.map(ix=>ys.map(iy=>Math.abs(iy-ix)).sort()[0]).sort()[0];

我不确定这是否真的更短或更好，只是另一种方法......

我只是过滤掉所有的 x 和 y 位置，然后计算它们之间的距离 (iy-ix) 并取最小值 (.sort()[0])

http://jsbin.com/nolohezape/edit?console

Answer 2

您可以通过测试实际差异是否更小来保留索引并优化最小值。

function getMinDistance(array, left, right) {
    var rightIndex, leftIndex, minDistance;
    array.forEach(function (a, i) {
        if (a === left && (leftIndex === undefined || leftIndex < i)) {
            leftIndex = i;
        }
        if (a === right && leftIndex !== undefined) {
            rightIndex = i;
        }
        if (leftIndex < rightIndex && (minDistance === undefined || minDistance > rightIndex - leftIndex)) {
            minDistance = rightIndex - leftIndex;
        }
    });
    return minDistance
}

console.log(getMinDistance(["2", "3", "5", "7", "1", "2", "3", "4", "8"], "5", "8"));
console.log(getMinDistance([3, 5, 4, 2, 6, 5, 6, 6, 5, 4, 8, 3], 3,  6));

Answer 3

 
function findMin(arr,a,b){
	var firstIndex = arr.indexOf(a);
  console.log(firstIndex);
  
  var lastIndex = arr.indexOf(b);
  console.log(lastIndex);
  
  var minDistance;
  
  if(firstIndex===lastIndex){
  		minDistance = 1;
  }
  if(firstIndex<lastIndex){
  		minDistance = lastIndex-firstIndex;
  }
	return minDistance;
}

 console.log(findMin([1,2,3,4,5,6],1,5));

Answer 4

假设数组可能非常大，我宁愿在迭代之前不要sort，因为这不是一个有效的策略。

下面的逻辑从左到右扫描数组，所以在每次迭代中，用?标记的数字与所有进行中的数字进行检查，直到找到最佳匹配，然后同样发生宽度下一个数字，直到计算出所有可能性并保存最佳（最低）结果 (minDis)。

 ?  -  -   -   -   -  ?
[50, 5, 75, 66, 32, 4, 58] // diff between 50 and each number past it (min is 8)

     ? -   -   -  ?  -
[50, 5, 75, 66, 32, 4, 58] // diff between 5 and each number past it (min is 1)

        ?  ?  -   -   -
[50, 5, 75, 66, 32, 4, 58] // diff between 75 and each number past it (min is 9)
...

在递归的每次迭代中，minDis 参数被发送到更深的级别，因此该级别的本地差异（for 循环）与 minDis 参数进行比较，如果差异较小，然后将其设置为“新”minDis 值：

var data = [50, 5, 75, 66, 32, 4, 58]; // assume a very large array

// find the minimum distance between two numbers
function findMinDistance(arr, minDis = Infinity, idx = 0){ 
  for( var numIdx = idx; numIdx < arr.length; numIdx++ ){
    var diff = Math.abs(arr[idx] - arr[numIdx+1])
    if( diff < minDis ) minDis = diff
    // no need to continue scanning, since "0" is the minimum possible
    if( minDis === 0 ) return 0
  }
  
  // scan from left to right, so each item is compared to the ones past it
  return idx < arr.length - 1 && minDis > 0 
    ? findMinDistance(arr, minDis, idx+1) 
    : minDis
}

console.log(`Min distance is: ${findMinDistance(data)}`)

一种非递归方法，类似于上述：

var data = [50, 5, 75, 66, 32, 4, 58]; // assume a very large array

// find the minimum distance between two numbers
function findMinDistance(arr){
  var minDis = Infinity, idx = 0, numIdx = idx
  
  for( ; numIdx < arr.length; numIdx++ ){
    var diff = Math.abs(arr[idx] - arr[numIdx+1])
    
    // if result is lower than minDis, save new minDis
    if( diff < minDis ) 
      minDis = diff 
    
    // "0" is the minimum so no need to continue
    if( minDis === 0 ) 
      return 0 
    
    // go to the next number and compare it to all from its right
    else if( numIdx == arr.length - 1 ) 
      numIdx = ++idx
  }

  return minDis
}

console.log(`min distance is: ${findMinDistance(data)}`)