由网友(执)分享简介：我试图确定最频繁的字符在具有字符作为其元素的矢量。 I am trying to determine the most frequent character in a vector that has chars as its elements. 我想这样做的：通过矢量循环并创建映射，其中一个关键是在载体中发现了一个...

我试图确定最频繁的字符在具有字符作为其元素的矢量。

I am trying to determine the most frequent character in a vector that has chars as its elements.

我想这样做的：

通过矢量循环并创建映射，其中一个关键是在载体中发现了一个独特的字符。对应的值将是炭的频率的整数计数。 在我穿过向量的所有元素都走了，地图上会 包含所有字符的频率。因此，我便要找到 哪个键具有最高值，因此，确定最 频繁的角色，在载体中。 looping through the vector and creating a map, where a key would be a unique char found in the vector. The corresponding value would be the integer count of the frequency of that char. After I have gone through all elements in the vector, the map will contain all character frequencies. Thus I will then have to find which key had the highest value and therefore determine the most frequent character in the vector.

这似乎很令人费解，虽然，所以我想知道，如果有人可以建议如果这种方法会被认为是可以接受的。在性能方面/良好的编码

This seems quite convoluted though, thus I was wondering if someone could suggest if this method would be considered 'acceptable' in terms of performance/good coding

能以更好的方式这样做？

推荐答案

如果您只使用普通的ASCII字符，可以使解决方案快一点 - 用地图来代替，用大小为256的数组和计数出现的字符与数组单元格给定的code'X'计数[X] 。这将删除从解决方案中的对数（256），从而将使其更快一点。我不认为其他更多关于这种算法的优化来实现。

If you are only using regular ascii characters, you can make the solution a bit faster - instead of using a map, use an array of size 256 and count the occurrences of the character with a given code 'x' in the array cell count[x]. This will remove an logarithm(256) from your solution and thus will make it a bit faster. I do not think much more can be done with respect to optimization of this algorithm.