Leetcode 713 Subarray Product Less Than K (子数组乘积大于K的个数) (双指针)

Posted 2021-03-03 Will

目录

• 问题描述
• 例子
• 方法

Leetcode 713

问题描述

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.

例子

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.

方法

** Solution Java **
** 7ms, beats 98.53% **
** 49.5MB, beats 100.00% **
class Solution {
    public int numSubarrayProductLessThanK(int[] nums, int k) {
        if (k == 0)
            return 0;
        int product = 1, res = 0, n = nums.length;
        for (int left = 0, right = 0; right < n; ++right){
            product *= nums[right];
            while (left <= right && product >= k) 
                product /= nums[left++];
            res += right - left + 1;
        }
        return res;
    }
}

** Solution Python3 **
** 1060ms, beats 36.99% **
** 16.7MB, beats 14.29% **
class Solution(object):
    def numSubarrayProductLessThanK(self, nums, k):
        if (k == 0) :
            return 0
        product, res, left = 1, 0, 0
        for right in range(len(nums)) :
            product *= nums[right]
            while (left <= right and k <= product) :
                product /= nums[left]
                left += 1
            res += right - left + 1
        return res