当前位置 博文首页 > 大树先生的博客:LeetCode-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: 8Explanation: 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:
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。
双指针遍历。
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;
}
}