leetcode 713. Subarray Product Less Than K

最新推荐文章于 2024-11-01 16:51:47 发布

原创最新推荐文章于 2024-11-01 16:51:47 发布 · 279 阅读

0 ·

CC 4.0 BY-SA版权

leetcode 专栏收录该内容

280 篇文章

订阅专栏

本文介绍了一种高效算法来计算一个整数数组中所有连续子数组的个数，其中这些子数组的乘积小于给定阈值K。通过使用双指针技巧，可以在O(n)的时间复杂度内解决问题。

713. Subarray Product Less Than K

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.

1、两个指针的思想。

i 指针代表以 i位开始的数字的种类

j 往后走，走到小于k的位置

( j - i )代表 i 开始的符合要求的个数。

2、当 i 指针追上 j 指针，说明这个数字已经比k大了，所以过不去。所以要判断一下。

class Solution {
public:
    int numSubarrayProductLessThanK(vector<int>& nums, int k)
    {
        if (k <= 1)  return 0;
        int ret = 0;
        int sum = 1;
        int j = 0;
        for (int i = 0; i < nums.size(); i++)
        {
            while (j < nums.size() && sum * nums[j] < k)
            {
                sum *= nums[j];
                j++;
            }
            ret += (j - i);
            if (i == j) //过不去坎了
            {
                sum = 1;
                j++;
            }
            else
                sum /= nums[i];
        }
        return ret;
    }
};