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]
这道题就是根据条件老老实实的做
1 /** 2 * Definition for ListNode 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { 7 * val = x; 8 * next = null; 9 * }10 * }11 */12 public class Solution {13 /**14 * @param hashTable: A list of The first node of linked list15 * @return: A list of The first node of linked list which have twice size16 */ 17 public ListNode[] rehashing(ListNode[] hashTable) {18 // write your code here19 int oldSize = hashTable.length;20 int newSize = oldSize * 2;21 if (hashTable==null || oldSize==0) return null;22 ListNode[] res = new ListNode[newSize];23 for (int i=0; i<oldSize; i++) {24 if (hashTable[i] != null) rehash(hashTable, res, i);25 }26 return res;27 }28 29 public void rehash(ListNode[] hashTable, ListNode[] res, int i) {30 int newSize = res.length;31 ListNode cur = hashTable[i];32 while (cur != null) {33 int val = cur.val;34 int newPos = val>=0? val%newSize : (val%newSize+newSize)%newSize;35 if (res[newPos] == null) res[newPos] = new ListNode(val);36 else {37 ListNode temp = res[newPos];38 while (temp.next != null) {39 temp = temp.next;40 }41 temp.next = new ListNode(val);42 }43 cur = cur.next;44 }45 }46 };
原标题:Lintcode: Rehashing
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