README.md

剑指 Offer II 009. 乘积小于 K 的子数组

题目描述

给定一个正整数数组 nums和整数 k ，请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。

示例 2:

输入: nums = [1,2,3], k = 0
输出: 0

 

提示: 

• 1 <= nums.length <= 3 * 104
• 1 <= nums[i] <= 1000
• 0 <= k <= 106

 

注意：本题与主站 713 题相同：https://leetcode.cn/problems/subarray-product-less-than-k/ 

解法

利用滑动窗口，我们能求出每个不同 right 结尾的合法子数组的个数

Python3

class Solution:
    def numSubarrayProductLessThanK(self, nums: List[int], k: int) -> int:
        n = len(nums)
        ans = 0
        sum = 1
        left, right = 0, 0
        while right < n:
            sum *= nums[right]
            right += 1
            while sum >= k and left < right:
                sum /= nums[left]
                left += 1
            ans += right - left
        return ans

Java

class Solution {
    public int numSubarrayProductLessThanK(int[] nums, int k) {
        int n = nums.length;
        int ans = 0;
        int sum = 1;
        int left = 0, right = 0;
        while (right < n) {
            sum *= nums[right++];
            while (sum >= k && left < right) {
                sum /= nums[left++];
            }
            ans += right - left;
        }
        return ans;
    }
}

Go

func numSubarrayProductLessThanK(nums []int, k int) int {
	n := len(nums)
	ans := 0
	sum := 1
	left, right := 0, 0
	for right < n {
		sum *= nums[right]
		right++
		for sum >= k && left < right {
			sum /= nums[left]
			left++
		}
		ans += right - left
	}
	return ans
}

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;
    }
};

...