2018-09-01 23:02:46

问题求解:

问题求解:

最开始的时候,一眼看过去就是一条 dp 嘛,保存每个数字结尾的长度和,最后求和就好,至于长度如何求,本题中需要用滑动窗口来维护。

很好的题目,将滑动窗口算法和动态规划巧妙的结合了起来。

    public int numSubarrayProductLessThanK(int[] nums, int k) {
if (k <= 1) return 0;
int res = 0;
int begin = 0;
int cur = 1;
for (int end = 0; end < nums.length; end++) {
cur *= nums[end];
while (cur >= k) {
cur /= nums[begin];
begin++;
}
res += end - begin + 1;
}
return res;
}

  

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