The size of the hash table is not determinate at the very beginning. If the total size of keys is too large (e.g. size >= capacity / 10), ...
The size of the hash table is not determinate at the very beginning. If the total size of keys is too large (e.g. size >= capacity / 10), we should double the size of the hash table and rehash every keys. Say you have a hash table looks like below:size=3, capacity=4[null, 21->9->null, 14->null, null]The hash function is:int hashcode(int key, int capacity) { return key % capacity;}here we have three numbers, 9, 14 and 21, where 21 and 9 share the same position as they all have the same hashcode 1 (21 % 4 = 9 % 4 = 1). We store them in the hash table by linked list.rehashing this hash table, double the capacity, you will get:size=3, capacity=8index: 0 1 2 3 4 5 6 7hash table: [null, 9, null, null, null, 21, 14, null]Given the original hash table, return the new hash table after rehashing .NoteFor negative integer in hash table, the position can be calculated as follow:In C++/Java, if you directly calculate -4 % 3 you will get -1. You can use function: a % b = (a % b + b) % b to make it is a non negative integer.In Python, you can directly use -1 % 3, you will get 2 automatically.ExampleGiven [null, 21->9->null, 14->null, null], return [null, 9->null, null, null, null, 21->null, 14->null, null]
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原标题：Lintcode: Rehashing
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