javascript - 找不到数组逆向算法问题的解决方案

Question

我有一个要求，我必须反转数组而不更改数组中存在的“#”索引，如下例所示:

• 数组 [18,-4,'#',0,8,'#',5] 应返回 [5, 8, "#", 0, -4, "#", 18]，这里的数字应该颠倒过来，排除'#'，同时保持相同的索引。

我尝试获得正确的输出，但似乎并非在所有情况下都是正确的:var arr = [18,-4,'#',0,8,'#',5];//给出的结果是正确的var arr1 = [18,-4,0,'#',8,'#',5];//结果不正确

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));

Answer 1

您可以简化函数并仅使用两个索引，即开始和结束，并检查索引处的值是否应保留，然后选择另一个索引进行交换。

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));