您好我被问到以下问题。
给定两个数组，即array1和array2。它们都按排序顺序包含数字。
Array1还包含-1等; array2中的数字与array1中的多个-1一样多。
示例如下，
array1 = [-1,-1,-1,-1,56,78,90,1200];
array2 = [1,4,5,1000]
我需要编写一个程序，将上面的数组合并为一个，它将按排序顺序包含两个数组中的数字，除了-1。
这是我的代码如下，
puzzle04([3,6,-1,11,15,-1,23,34,-1,42],[1,12,28]);
puzzle04([3,6,-1,11,15,-1,23,34,-1,42],[7,19,38]);
puzzle04([3,6,11,15,32,34,42,-1,-1,-1,-1],[1,10,17,56]);
puzzle04([-1,-1,-1,-1,3,6,11,15,32,34,42],[1,10,17,56]);
puzzle04([-1,-1,-1,-1,3,6,11,15,32,34,42],[56,78,90,100]);
puzzle04([12,34,65,-1,71,85,90,-1,101,120,-1,200],[24,37,94]);
puzzle04([3,6,-1,11,15,-1,32,34,-1,42,-1],[1,10,17,56]);
puzzle04([-1,-1,-1,56,78,90,112],[1,4,5]);
puzzle04([-1,-1,-1,-1,56,78,90,112],[1,4,5,1000]);
puzzle04([-1,-1,-1,-1,56,78,90,1200],[1,4,5,1000]);
function puzzle04(array1,array2){
var outputArray = [],
array1Counter = 0, // counter for array1
array2Counter = 0, // counter for array2
isArray2NumPlaced = false, // has number from array2 found its position in output array ?
areAllArray2NumsFilled = false; // is number pushed in output array
// iterating through array2 loop
for(array2Counter = 0; array2Counter < array2.length; array2Counter++){
// iterating through array1 loop
for(; (isArray2NumPlaced === false); array1Counter++){
// -1 encountered in array1
if(array1[array1Counter] === -1){
continue;
// if array1 number is less than array2 number
// then push array1 number in ouput array
}else if(array1[array1Counter] < array2[array2Counter]){
outputArray.push(array1[array1Counter]);
}else{ // array2 number is less then array1 number
// add array2 number in output array until
// all array2 numbers are not added in output array.
if(areAllArray2NumsFilled === false){
outputArray.push(array2[array2Counter]);
}
// is array2 number pushed in output array ?
isArray2NumPlaced = true;
}// end of if-else
// if all the array2 numbers are added in output array
// but still array1 numbers are left to be added
if(isArray2NumPlaced === true
&& array2Counter === (array2.length - 1)
&& array1Counter <= (array1.length - 1)){
outputArray.push(array1[array1Counter]);
// set the below flag to false so that,
// array1 loop can iterate
isArray2NumPlaced = false;
// all the numbers of array2 are entered in output array
areAllArray2NumsFilled = true;
}// end of if
}// array1 for-loops ends
array1Counter--;
isArray2NumPlaced = false;
}// array2 for-loops ends
console.log("final ",outputArray);
}
以上代码的输出如下，
final [ 1, 3, 6, 11, 12, 15, 23, 28, 34, 42 ]
final [ 3, 6, 7, 11, 15, 19, 23, 34, 38, 42 ]
final [ 1, 3, 6, 10, 11, 15, 17, 32, 34, 42, 56 ]
final [ 1, 3, 6, 10, 11, 15, 17, 32, 34, 42, 56 ]
final [ 3, 6, 11, 15, 32, 34, 42, 56, 78, 90, 100 ]
final [ 12, 24, 34, 37, 65, 71, 85, 90, 94, 101, 120, 200 ]
final [ 1, 3, 6, 10, 11, 15, 17, 32, 34, 42, 56 ]
final [ 1, 4, 5, 56, 78, 90, 112 ]
final [ 1, 4, 5, 56, 78, 90, 112, 1000 ]
final [ 1, 4, 5, 56, 78, 90, 1000, 1200 ]
当我向审稿人展示我的代码时，他说我使用了太多的布尔变量，代码可以更加简单。
我尽力即兴发挥，但没有任何线索。
您能否建议我解决上述问题的更好方法
注意：不能使用任何现成的排序方法或预先编写的api来解决上述练习。
答案 0 :(得分：3)
您所要做的就是逐步浏览两个数组，取两个值中较小的一个，然后将其添加到输出列表中。一旦添加了所有一个数组，另一个数组的其余部分都会变大，并且可以一次性添加。
function merge(x, y) {
var i = 0;
var j = 0;
var result = [];
while (i < x.length && j < y.length) {
// Skip negative numbers
if (x[i] === -1) {
x += 1;
continue;
}
if (y[j] === -1) {
y += 1;
continue;
}
// Take the smaller of the two values, and add it to the output.
// Note: the index (i or j) is only incremented when we use the corresponding value
if (x[i] <= y[j]) {
result.push(x[i]);
i += 1;
} else {
result.push(y[j]);
j += 1;
}
}
// At this point, we have reached the end of one of the two arrays. The remainder of
// the other array is all larger than what is currently in the output array
while (i < x.length) {
result.push(x[i]);
i += 1;
}
while (j < y.length) {
result.push(y[j]);
j += 1;
}
return result;
}
答案 1 :(得分：1)
function merge(array1, array2)
{
int i = 0, j = 0;
array3; //answer
while ( i<array1.size() and j<array2.size() ){
if (array1[i] == -1 ) //ignore -1
{ ++i; continue; }
if (array2[j] == -1 ) //ignore -1
{ ++j; continue; }
if ( array1[i] < array2[j] ) { //choose the minor
array3.push( array1[i] );
++i;
}
else {
array3.push( array2[j] );
++j;
}
}
//if some array has elements still
while (i < array1.size() )
array3.push(array1[i]);
while (j < array2.size() )
array3.push(array1[j]);
return array3
}
答案 2 :(得分：1)
要将2个已排序的阵列拼接在一起，您可以按如下所示进行拼接。要添加省略特定值的功能，可以在合并数组之前使用 var arglist = new Dictionary<string, object>() {
{ "1",1 } , {"2",2 }, {"3",new object() }
};
string testval = CustomerInfo(arglist);
来消除这些元素。
.filter()
答案 3 :(得分：1)
function merge(array1, array2){
var array = [];
var i = 0;
var j = 0;
while(i < array1.length || j < array2.length){
if(array1[i] === -1){
i++;
continue;
}
if(array1[i] && (!array2[j] || array1[i] <= array2[j])){
array.push(array1[i]);
i++;
} else if(array2[j] && (!array1[i] || array2[j] <= array1[i])){
array.push(array2[j]);
j++;
}
}
return array;
}
答案 4 :(得分：1)
试试这个
function puzzle04(arr1, arr2)
{
var result = [];
var final = [];
var arr_concat = arr1.concat(arr2).filter(function (value) {
return value != -1;
});
arr_concat.forEach(function(x) {
if(!result[x]) {
result[x] = [];
result[x].push('super sort');
}
else {
result[x].push('super sort');
}
});
result.forEach(function(k, v) {
k.forEach(function(x) {
final.push(v);
});
});
console.log(final);
}
答案 5 :(得分：1)
您所描述的是mergeSort：
function puzzle04(array1, array2) {
return mergeSortWithoutMinusOnes(array1.concat(array2))
}
function mergeSortWithoutMinusOnes(items){
if (items.length < 2) {
return items;
}
var middle = Math.floor(items.length / 2),
left = items.slice(0, middle),
right = items.slice(middle);
return mergeWithoutMinusOnes(mergeSortWithoutMinusOnes(left), mergeSortWithoutMinusOnes(right));
}
function mergeWithoutMinusOnes(left, right){
var result = [],
il = 0,
ir = 0;
while (il < left.length && ir < right.length){
if (left[il] < right[ir]){
if (left[il] === -1) {
il++;
} else {
result.push(left[il++]);
}
} else {
if (right[ir] === -1) {
ir++
} else {
result.push(right[ir++]);
}
}
}
return result.concat(left.slice(il)).concat(right.slice(ir));
}
答案 6 :(得分：1)
考虑您的特殊-1排除条件;您可以选择将多个已排序的数组合并为一个，如下所示;
function merger(a,b,_,__,r = []){
return !a.length ? !b.length ? r
: r.concat(b.filter(n => n > -1))
: !b.length ? r.concat(a.filter(n => n > -1))
: a[0] > -1 ? b[0] > -1 ? a[0] > b[0] ? (r.push(b[0]), merger(a,b.slice(1),_,__,r))
: (r.push(a[0]), merger(a.slice(1),b,_,__,r))
: merger(a,b.slice(1),_,__,r)
: merger(a.slice(1),b,_,__,r);
}
var arrs = [[3,6,-1,11,15,-1,23,34,-1,42],[1,12,28],[3,6,-1,11,15,-1,23,34,-1,42],[7,19,38],[3,6,11,15,32,34,42,-1,-1,-1,-1],[1,10,17,56],[-1,-1,-1,-1,3,6,11,15,32,34,42],[1,10,17,56],[-1,-1,-1,-1,3,6,11,15,32,34,42],[56,78,90,100],[12,34,65,-1,71,85,90,-1,101,120,-1,200],[24,37,94],[3,6,-1,11,15,-1,32,34,-1,42,-1],[1,10,17,56],[-1,-1,-1,56,78,90,112],[1,4,5],[-1,-1,-1,-1,56,78,90,112],[1,4,5,1000],[-1,-1,-1,-1,56,78,90,1200],[1,4,5,1000]],
res = arrs.reduce(merger);
console.log(JSON.stringify(res));
答案 7 :(得分：0)
您可以在一个循环中迭代arrayOne，然后arrayTwo。
这个想法是将目标索引与实际索引分开。第一个循环忽略-1并保留目标索引。
如果实际值大于arrayTwo的第一个值，则会交换这两个值，并通过迭代并使用更大的值进行交换来在arrayTwo中进行排序。
然后，将实际项目分配给目标索引。
两个指数都会增加。
最后，所有arrayTwo项都会添加到arrayOne。
function puzzle04(arrayOne, arrayTwo) {
var i = 0, j, l = 0;
while (i < arrayOne.length) {
if (arrayOne[i] === -1) {
i++;
continue;
}
if (arrayTwo[0] < arrayOne[i]) {
[arrayOne[i], arrayTwo[0]] = [arrayTwo[0], arrayOne[i]];
j = 0;
while (arrayTwo[j] > arrayTwo[j + 1]) {
[arrayTwo[j], arrayTwo[j + 1]] = [arrayTwo[j + 1], arrayTwo[j]];
j++;
}
}
arrayOne[l++] = arrayOne[i++];
}
j = 0;
while (l < arrayOne.length) {
arrayOne[l++] = arrayTwo[j++];
}
return arrayOne;
}
console.log(puzzle04([3, 6, -1, 11, 15, -1, 23, 34, -1, 42], [1, 12, 28]));
console.log(puzzle04([3, 6, -1, 11, 15, -1, 23, 34, -1, 42], [7, 19, 38]));
console.log(puzzle04([3, 6, 11, 15, 32, 34, 42, -1, -1, -1, -1], [1, 10, 17, 56]));
console.log(puzzle04([-1, -1, -1, -1, 3, 6, 11, 15, 32, 34, 42], [1, 10, 17, 56]));
console.log(puzzle04([-1, -1, -1, -1, 3, 6, 11, 15, 32, 34, 42], [56, 78, 90, 100]));
console.log(puzzle04([12, 34, 65, -1, 71, 85, 90, -1, 101, 120, -1, 200], [24, 37, 94]));
console.log(puzzle04([3, 6, -1, 11, 15, -1, 32, 34, -1, 42, -1], [1, 10, 17, 56]));
console.log(puzzle04([-1, -1, -1, 56, 78, 90, 112], [1, 4, 5]));
console.log(puzzle04([-1, -1, -1, -1, 56, 78, 90, 112], [1, 4, 5, 1000]));
console.log(puzzle04([-1, -1, -1, -1, 56, 78, 90, 1200], [1, 4, 5, 1000]));.as-console-wrapper { max-height: 100% !important; top: 0; }