乘积小于K的子数组

https://leetcode-cn.com/problems/ZVAVXX/

class Solution {
    public int numSubarrayProductLessThanK(int[] nums, int k) {
        int left = 0, ans = 0;
        int sum = 1;
	
		//
        for (int i = 0; i < nums.length; i ++) {
        	//当前子数组乘积
            sum *= nums[i];
			
			//如果乘积不小于k，那么在i左边，每次减少一个数组元素的乘积，来找出可能的子数组
            while (left <= i && sum >= k) {
                sum /= nums[left ++];
            }
			
			//此时的left<=i且[left, i]的乘积小于k
            if (left <= i) {
                ans += i - left + 1;
            }
        }
        return ans;
    }
}

标签：小于,乘积,nums,int,sum,数组,ans,left
来源： https://blog.csdn.net/qq_43806430/article/details/122413244