s[零]31^(n-1) + s[1]31^(n-2) + . + s[n-1] 用这种算法做hash

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Jun 13, 2016 pm 01:19 PM

hashhashcodejava

s[0]*31^(n-1) + s[1]*31^(n-2) + ... + s[n-1] 用这种算法做hash
s[0]*31^(n-1) + s[1]*31^(n-2) + ... + s[n-1]用这种算法做hash

怎么样？

怎么检测冲突？

当我是一段定长的字符串时候比如说 10个字符的时候

冲突情况是怎么样的？

据说这个算法是 java的hashcode中的不知道是不是？我就发java版吧

------解决方案--------------------
海量数据的 hash 冲突是必然存在的，一般是在出现冲突时，采用顺序列表
如果你没有预留出现冲突时的存储空间，多半无解

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