将两个排序的数组合并为一个
嗨我被问到以下问题。
给定两个数组，即array1和array2。它们都按排序顺序包含数字。
Given two arrays i.e. array1 and array2. Both of them contain numbers in sorted order.
Array1还包含-1，如; array2中的数字与array1中的多个-1一样多。
Array1 also contains -1 such as; as many numbers in array2 that many -1's in array1.
示例如下，
array1 = [-1,-1,-1,-1,56,78,90,1200];
array2 = [1,4,5,1000]
我需要写一个程序它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。
I need to write a program which merge the above arrays in one, which will contain numbers from both the arrays in sorted order, except -1.
这是我的代码，如下所示，
This is my code as follows,
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);
}
上述代码的输出如下，
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，而且代码可以更加简单。
When I showed my code to reviewer, he said I used too much boolean variables and the code can be lot more simpler.
我尽力做到即兴，但没有我有任何线索。
I tried my best to improvise, but didn't got any clue.
你能否建议我解决上述问题的更好方法
Can you please suggest me any better way to solve above problem
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。
Note: can't use any ready-made sorting methods or pre written api to solve above excercise.
您所要做的就是逐步浏览两个数组，取两个值中的较小值，然后将其添加到输出中名单。一旦你添加了所有的一个数组，其他数组的其余部分都会更大，并且可以一次性添加。
All you have to do is step through both arrays, take the smaller of the two values, and add it to the output list. Once you have added all of one array, the remainder of the other array is all larger, and can be added in one go.
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;
}