[Leetcode] 713. Subarray Product Less Than K 解题报告

题目：

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

.
思路：

Two pointers问题：我们每次增加一个nums[end]，然后从start开始，一旦发现[start, end]区间内的乘积大于k，就增加start。当[start, end]区间内的乘积小于k的时候，就会有(end - start + 1)个子区间符合条件。最后返回总的个数即可。算法的时间复杂度是O(n)，空间复杂度是O(1)。

代码：

class Solution {
public:
int numSubarrayProductLessThanK(vector<int>& nums, int k) {
int count = 0, start = 0, product = 1;
for (int end = 0; end < nums.size(); ++end) {
product *= nums[end];
while (start <= end && product >= k) {
product /= nums[start];
++start;
}
count += (end - start + 1);
}
return count;
}
};