javascript - 合并排序 : merge function array index zero returning multiple indexes

Question

我想用 Javascript 实现归并排序作为一种学习体验。我有函数 mergeSort(unsortedArray)，它接受一个未排序的数组并使用合并排序策略对其进行排序。 mergeSort() 调用 merge(leftArray,rightArray)，它将两个已排序的数组合并在一起，生成一个已排序的数组。

我认为问题出在 merge() 函数上。在数组上调用 mergeSort 时:[8,8,7,5,4,6,3,2,1,5,9,8,7,6,5,4,2,3,6,5,4, 8] 我得到结果:[1,4,2,3,5,5,9,6,7,8,8]。据我所知，问题的根源是在 merge() 函数中，在比较 leftArray[0] 和 rightArray[0] 时，rightArray[0] 有时会返回多个值，而不仅仅是第一个索引.在我的例子中，它是用 2,3 和 5,9 来做的。所以代码运行时，有时rightArray[0] = 2,3，2,3拼接出数组后rightArray[0]=5,9。以下是出现此问题时 merge() 中发生的情况:

第一步

左数组:[4,5,6,7,8,8]
rightArray:[1,2,3,5,9]
结果:[]

第二步

左数组[4,5,6,7,8,8]
rightArray[2,3,5,9]
结果:[1]

第三步

(不正确的索引...array[0] 返回两个值)
左数组[0]=4
rightArray[0]=2,3

左数组[5,6,7,8,8]
rightArray[2,3,5,9]
结果[1,4]

第四步

(不正确的索引...array[0] 返回两个值)
左数组[0]=5
rightArray[0]=2,3

左数组[5,6,7,8,8]
rightArray[5,9]
结果[1,4,2,3]

...array[0] 索引再次搞砸，接下来返回 rightArray[0] = 5,9。奇怪的是，如果我在 leftArray=[4,5,6,7,8,8] 和 rightArray[1,2,3,5,9] 上独立于 mergeSort() 调用我的 merge() 函数，它工作正常并且返回正确的结果，没有奇怪的索引行为。

//Implement Merge Sort...
    function mergeSort(unsortedArray) {
        var leftArray = [];
        var rightArray = [];
        var result = [];
        
        //Base Case of one element
        if(unsortedArray.length <= 1){
            //alert("Array is size 1 and value: " + unsortedArray);
            return unsortedArray;
        }
        else{
            var halfwayPoint = Math.round(unsortedArray.length/2);
            
            //Sepertate unsortedArray into a left and right array
            for(var i = 0; i < halfwayPoint; i++){
                leftArray.push(unsortedArray[i]);
                //alert("leftArray: "+ leftArray + " index i = " + i);
            }
            for(var i = halfwayPoint; i < unsortedArray.length; i++){
                rightArray.push(unsortedArray[i]);
                //alert("rightArray" + rightArray + " index i = " + i);
            }
            //alert("leftArray: " + leftArray + " rightArray: " + rightArray);
            leftArray = mergeSort(leftArray);
            rightArray = mergeSort(rightArray);
            //alert("Arrays before merge = leftArray: " + leftArray + " rightArray: " + rightArray);
            result = merge(leftArray, rightArray);
            //alert("result: " + result);
        }
        return result;
    }
    
    //Helper function Merge for MergeSort
    function merge(leftArray, rightArray)
    {
        var result = [];
        while(leftArray.length > 0 && rightArray.length > 0){
            //compare first items of both lists
            //alert("top of while loop");
            //alert("leftArray[0] = " + leftArray[0] + " rightArray[0] = " + rightArray[0]);
            if(leftArray[0] >= rightArray[0]){
                result.push(rightArray[0]);
                //alert("result after push rightArray[0] " + result + " and rightArray before splice: "+ rightArray);
                rightArray.splice(0,1);
                //alert("rightArray after splce: " + rightArray);
            }
            else{
                result.push(leftArray[0]);
                //alert("result after push leftArray[0] " + result + " and leftArray before splice: "+ leftArray);
                leftArray.splice(0,1);
                //alert("leftArray after splce: " + leftArray);
            }
        }
        //alert("before leftArray add");
        if(leftArray.length > 0){
            //alert("went into left array > 0 leftArray: " + leftArray);
            result.push(leftArray);
        }
        //alert("before rightArray add");
        if(rightArray.length > 0){
            //alert("went into right array > 0 rightArray: " + rightArray);
            result.push(rightArray);
        }
        //alert("result within merge function: " + result);
        return result;
    }
    //Test Case
    var unsortedArray = [8,8,7,5,4,6,3,2,1,5,9,8,7,6,5,4,2,3,6,5,4,8];
    var sortedArray = mergeSort(unsortedArray);
    lert(sortedArray);
  
    //Problem is when Merge sort has left array and right array described below
    //the merge function will yield proper result on left array and right array
    //if called directly as it is below, however when merge is called through
    //mergeSort with leftArray and rightArray as described below it yields
    // improperResult below
    var leftArray = [4,5,6,7,8,8];
    var rightArray = [1,2,3,5,9];
    var improperResult= [1,4,2,3,5,5,9,6,7,8,8];
    var resultAct = merge(leftArray,rightArray);
    alert(resultAct);

<h1>MergeSort Problem</h1>

Answer 1

您需要使用 Array.prototype.concat()而不是 .push() 来连接 2 个 arrays.

.concat 合并 2 个(或更多)数组并返回一个新的 array，而 push 只是将目标放在 array 的末尾>，它不会为您连接数组。

如果您记录原始结果而不是警报，您会看到

[1, 2, 3, 4, 4, Array[2], 5, Array1, Array[2], Array1, Array[2], Array[4]]

很明显，您只是将数组插入了结果。

所以在你的

if(leftArray.length > 0){
    result.push(leftArray);
}
if(rightArray.length > 0){
    result.push(rightArray);
}

你应该写信给:

if(leftArray.length > 0){
    result = result.concat(leftArray);
}
if(rightArray.length > 0){
  result = result.concat(rightArray);
}

    function mergeSort(unsortedArray) {
        var leftArray = [];
        var rightArray = [];
        var result = [];
        
        //Base Case of one element
        if(unsortedArray.length <= 1){
            return unsortedArray;
        }
        else{
            var halfwayPoint = Math.round(unsortedArray.length/2);
            
            //Sepertate unsortedArray into a left and right array
            for(var i = 0; i < halfwayPoint; i++){
                leftArray.push(unsortedArray[i]);
            }
            for(var i = halfwayPoint; i < unsortedArray.length; i++){
                rightArray.push(unsortedArray[i]);
            }

            leftArray = mergeSort(leftArray);
            rightArray = mergeSort(rightArray);
          
            result = merge(leftArray, rightArray);
        }
        return result;
    }
    
    //Helper function Merge for MergeSort
    function merge(leftArray, rightArray)
    {
        var result = [];
      
        while(leftArray.length > 0 && rightArray.length > 0){
            //compare first items of both lists
            if(leftArray[0] >= rightArray[0]){
                result.push(rightArray[0]);
                rightArray.splice(0,1);
            }
            else{
                result.push(leftArray[0]);
                leftArray.splice(0,1);
            }
        }
      
        if(leftArray.length > 0){
            result = result.concat(leftArray);
        }
        if(rightArray.length > 0){
          result = result.concat(rightArray);
        }

        return result;
    }
    //Test Case
    var unsortedArray = [8,8,7,5,4,6,3,2,1,5,9,8,7,6,5,4,2,3,6,5,4,8];
    var sortedArray = mergeSort(unsortedArray);
    alert(sortedArray);
  
    //Problem is when Merge sort has left array and right array described below
    //the merge function will yield proper result on left array and right array
    //if called directly as it is below, however when merge is called through
    //mergeSort with leftArray and rightArray as described below it yields
    // improperResult below
    var leftArray = [4,5,6,7,8,8];
    var rightArray = [1,2,3,5,9];
    var improperResult= [1,4,2,3,5,5,9,6,7,8,8];
    var resultAct = merge(leftArray,rightArray);
    alert(resultAct);

<h1>MergeSort Problem</h1>