我是靠谱客的博主 甜美中心,这篇文章主要介绍【leetcode每日刷题】【尺取法】713. Subarray Product Less Than K,现在分享给大家,希望可以做个参考。

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.
class num713 {
public int numSubarrayProductLessThanK(int[] nums, int k) {
if(k <= 1) return 0;
int prod = 1, ans = 0, left = 0;
for(int right=0; right< nums.length; right++){
prod *= nums[right];
while(prod >= k) prod /= nums[left++];
ans += right-left+1;
}
return ans;
}
public static void main(String[] args) {
num713 s = new num713();
int[] nums = {10, 5, 2, 6};
int
k = 100;
int result = s.numSubarrayProductLessThanK(nums, k);
System.out.println(result);
}
}

 

最后

以上就是甜美中心最近收集整理的关于【leetcode每日刷题】【尺取法】713. Subarray Product Less Than K的全部内容,更多相关【leetcode每日刷题】【尺取法】713.内容请搜索靠谱客的其他文章。

本图文内容来源于网友提供,作为学习参考使用,或来自网络收集整理,版权属于原作者所有。
点赞(63)

评论列表共有 0 条评论

立即
投稿
返回
顶部