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English Version

题目描述

给定一个正整数数组 nums

找出该数组内乘积小于 k 的连续的子数组的个数。

示例 1:

输入: nums = [10,5,2,6], k = 100
输出: 8
解释: 8个乘积小于100的子数组分别为: [10], [5], [2], [6], [10,5], [5,2], [2,6], [5,2,6]。
需要注意的是 [10,5,2] 并不是乘积小于100的子数组。

说明:

  • 0 < nums.length <= 50000
  • 0 < nums[i] < 1000
  • 0 <= k < 10^6

解法

C++

class Solution {
public:
    int numSubarrayProductLessThanK(vector<int>& nums, int k) {
        int left = 0, right;
        long mul = 1;
        int count = 0;

        for (right = 0; right < nums.size(); right++) {
            mul *= nums[right];

            while(left <= right && mul >= k) {
                mul /= nums[left++];
            }

            count += right >= left? right - left + 1: 0;
        }

        return count;
    }
};

Python3

Java

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