Can't find a solution for array reverse algorithm problem

I have a requirement where I have to reverse an array without changing the index of '#' presents in an array, like below example:

• Array [18,-4,'#',0,8,'#',5] should return [5, 8, "#", 0, -4, "#", 18], here numbers should be reversed, excluding '#' while keeping the same index.

I have tried to get the correct output, but it doesn't seem to be correct in all scenarios:
var arr = [18,-4,'#',0,8,'#',5]; // giving result is correct
var arr1 = [18,-4,0,'#',8,'#',5]; // result is not correct

<!-- begin snippet: js hide: false console: true babel: false -->

<!-- language: lang-js -->

var reverse = function(numbers, start, end){
	var temp = numbers[start];
  numbers[start] = numbers[end];
  numbers[end] = temp;
}
var flip = function(numbers) {
var start = 0;
var end = numbers.length-1;
	for(var i=0;i<parseInt(numbers.length/2);i++) {
  	if(numbers[i] === '#') {
    	start = i+1;
      end = numbers.length - i - i;
      reverse(numbers, start, end);
    } else if (numbers[numbers.length - i - 1] === '#') {
    	start = i;
      end = numbers.length - i - 2;
      reverse(numbers, start, end);
    } else {
    	reverse(numbers, start, end);
    }
  }
  return numbers;
}

var arr = [18,-4,'#',0,8,'#',5];
var arr1 = [18,-4,0,'#',8,'#',5];
console.log(flip(arr));
console.log(flip(arr1));

<!-- end snippet -->

You could simplify the function and use only two indices, the start and end and check if the value at the indices should stay, then choose another index for swapping.

<!-- begin snippet: js hide: false console: true babel: false -->
<!-- language: lang-js -->
const
swap = (array, a, b) => [array[a], array[b]] = [array[b], array[a]],
flip = numbers => {
var start = 0,
end = numbers.length - 1;

        while (start < end) {
            if (numbers[start] === '#') {
                start++;
                continue;
            }
            if (numbers[end] === '#') {
                end--;
                continue;
            }
            swap(numbers, start++, end--);
        }
        return numbers;
    },
    array1 = [18, -4, '#', 0, 8, '#', 5],
    array2 = [18, -4, 0, '#', 8, '#', 5];

console.log(...flip(array1));
console.log(...flip(array2));

<!-- end snippet -->

The trivial approach would be to remove all '#''s, reverse the array using the built in [].reverse method, and then re-insert the '#''s:

<!-- begin snippet: js hide: false console: true babel: false -->

<!-- language: lang-js -->

let flip = numbers => {
  let removed = numbers.reduce((r, v, i) =>
          v === '#' ? r.concat(i) : r
      , []);

  let reversed = numbers.filter(v => v !== '#').reverse();

  removed.forEach(i => reversed.splice(i, 0, '#'));

  return reversed;
};

let arr = [18, -4, '#', 0, 8, '#', 5];
let arr1 = [18, -4, 0, '#', 8, '#', 5];
console.log(flip(arr));
console.log(flip(arr1));

<!-- end snippet -->

You can try this:

var numbers = arr.filter(a => a !== '#')
var revArr = [];
arr.forEach((currentValue) => {
    if(currentValue !== "#") {
        revArr.push(numbers.pop());
    } else {
        revArr.push("#");
    }
});

You can base your algorithm on two basic array (array of reversed numbers and and an array with "#" saved positions)

<!-- begin snippet: js hide: false console: true babel: false -->

<!-- language: lang-js -->

const array = [5, 8, "#", 0, -4, "#", 18];

function flip(array) {

  const arrayNumbers = array.filter((el, index) => el !== "#").reverse();

  var counter = 0;

  return array.map(el => el === "#").map(el => {
    if (!el) {
      let num = arrayNumbers
0
+
网站访问量
;
      counter = counter + 1;
      return num;
    } else {
      return "#"
    }
  })
}

console.log(flip(array));

<!-- end snippet -->

var arr = [18,-4,'#',0,8,'#',5]
var stack = []

for (i=0 ; i<arr.length; i++) {
	if (arr[i] == '#') continue;
	stack.push(arr[i]);
}

for (i=0 ; i<arr.length; i++) {
	if (arr[i] != '#') {
		arr[i] = stack.pop();
	}
}

console.log(arr)

The above code should solve your problem.
<br>
The implementation uses stack where we keep inserting elements into stack untill we see a '#' and skip it.
While creating the output array we refer the original array for '#' index and stack for reverse index.

You can do it like this.

		int end = v.length - 1;
		int start = 0;
		for (int i = 0; i < v.length >> 1; i++) {
			if (v[start].equals("#")) {
				start++;
				continue;
			}
			if (v[end].equals("#")) {
				end--;
				continue;
			}
			Object temp = v[end];
			v[end] = v[start];
			v[start] = temp;
			end--;
			start++;

		}
		System.out.println(Arrays.toString(v));

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