Subarray Product Less Than K (M)

题目

Your are given an array of positive integers nums.

Count and print the number of (contiguous) subarrays where the product of all the elements in the subarray is less than k.

Example 1:

Input: nums = [10, 5, 2, 6], k = 100
Output: 8
Explanation: The 8 subarrays that have product less than 100 are: [10], [5], [2], [6], [10, 5], [5, 2], [2, 6], [5, 2, 6].
Note that [10, 5, 2] is not included as the product of 100 is not strictly less than k.

Note:

0 < nums.length <= 50000.

0 < nums[i] < 1000.

0 <= k < 10^6.

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题意

找到所有的连续子数组，使其积小于指定值。

思路

遍历数组所有元素作为区间右端点；对于每个右端点，找到最左侧的元素，使构成的区间的积小于指定值；累加以右端点为终点的子区间的数量。

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代码实现

Java

    public int numSubarrayProductLessThanK(int[] nums, int k) {
        int count = 0;
        int left = 0, product = 1;
        for (int right = 0; right < nums.length; right++) {
            product *= nums[right];
            while (product >= k && left <= right) {
                product /= nums[left++];
            }
            count += right - left + 1;
        }
        return count;
    }