您好，有人问我以下问题。
给定两个数组，即 array1 和 array2。它们都包含按排序顺序排列的数字。
Array1 还包含 -1 例如； array2 中的数字与 array1 中的 -1 一样多。
例子如下，
array1 = [-1,-1,-1,-1,56,78,90,1200];
array2 = [1,4,5,1000]
我需要编写一个程序，将上述数组合并为一个，其中将按排序顺序包含两个数组中的数字，-1 除外。
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);
}
上面代码的输出结果如下，
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 ]
当我向审阅者展示我的代码时，他说我使用了太多的 bool 变量，代码可以更简单。
我尽力即兴发挥，但没有得到任何线索。
能否请你建议我任何更好的方法来解决上述问题
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。
最佳答案
您所要做的就是遍历两个数组，取两个值中较小的一个，并将其添加到输出列表中。一旦你添加了一个数组的所有内容，另一个数组的剩余部分就更大了，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;
}
关于javascript - 将两个排序数组合并为一个，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/42817212/