如何在线性时间内找到一个数组中所有可能的子数组的乘积？

function findSubarrayProducts(nums) {
  const n = nums.length;
  const max_product = new Array(n);
  const min_product = new Array(n);
  let max_value = Number.MIN_SAFE_INTEGER;
  let start = -1;
  let end = -1;

  max_product[0] = nums[0];
  min_product[0] = nums[0];

  for (let i = 1; i < n; i++) {
    if (nums[i] > 0) {
      max_product[i] = Math.max(max_product[i - 1] * nums[i], nums[i]);
      min_product[i] = Math.min(min_product[i - 1] * nums[i], nums[i]);
    } else if (nums[i] < 0) {
      max_product[i] = Math.max(min_product[i - 1] * nums[i], nums[i]);
      min_product[i] = Math.min(max_product[i - 1] * nums[i], nums[i]);
    } else {
      max_product[i] = 0;
      min_product[i] = 0;
    }

    if (max_product[i] > max_value) {
      max_value = max_product[i];
      start = i;
      end = i;
    }
  }

  // 寻找最大乘积的子数组的起始位置和结束位置
  for (let i = start - 1; i >= 0; i--) {
    if (nums[i] === 0) {
      break;
    }
    start = i;
  }

  for (let i = end + 1; i < n; i++) {
    if (nums[i] === 0) {
      break;
    }
    end = i;
  }

  // 根据起始位置和结束位置找到所有可能的子数组，并计算乘积
  const subarray_products = [];
  for (let i = start; i <= end; i++) {
    for (let j = i; j <= end; j++) {
      const subarray = nums.slice(i, j + 1);
      const product = subarray.reduce((a, b) => a * b);
      subarray_products.push({
        subarray: subarray,
        product: product
      });
    }
  }

  return subarray_products;
}

const nums = [1, 2, 3, 4];
const subarray_products = findSubarrayProducts(nums);
console.log(subarray_products);