LeetCode-643. Maximum Average Subarray I
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Description:
Example 1:
Note:
Solution:
算法分析:
这道题也不难,注意在i>k的时候,sum是将各个项进行累加,当i>k之后才会用sum+=nums[i]-nums[i-k],然后就是循环里要想进行最大值比较,原因很简单,因为当最初的sum累加完成之后就需要一次比较。不然就会遗漏这次的sum。
程序分析:
这个题还有一个小地方卡了一下,那就是计算小数。对于这道题,有两种方法:
int a,b;double res=(double)a/b;
要么:
double a,b;double res=a/b;
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