LeetCode-713：Subarray Product Less Than K (乘积小于K的子数组个数) -- medium

Question

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.

问题解析：

给定数组，和整数K，计数数组中所有满足元素乘积小于K的连续子数组的个数。

Answer

Solution 1：

双指针遍历。

该题目主要的解题思路是寻找每步相同的计数规律。

按照一般的思维，要想记录连续子数组需要从左向右遍历数组。

这里最重要的就是处理乘积大于K的情况，需要丢弃最前面的元素，那么说明需要有一个指针记录最前面元素的位置。

在不大于K的情况下，则累加每次加入新元素后子数组的个数。

class Solution {
public int numSubarrayProductLessThanK(int[] nums, int k) {
if ( k <= 1) return 0;

int n = nums.length;
long p = 1l;
int i = 0, total = 0;
for (int j = 0; j < n; j++){
p *= nums[j];
while (p >= k){
p /= nums[i];
i++;
}
total += (j - i + 1);
}

return total;
}
}

时间复杂度：O(n)，空间复杂度：O(1)