哈斯一般用开放地址和拉链法来实现，开放地址的最坏情况要比拉链的可能性大，我的理解是开放地址形成冲突都是往后退然后放，
有可能形成要遍历全部数据才能找到要找的数据，而拉链的情况下一般来说，不会导致所有数据都同一条链表上。
至于平均复杂度，我不是很会分析，谁懂告诉我吧 ：)
开放地址的源码：http://blog.csdn.net/kangquan2008/article/details/6772103
拉链法源码：
/******************************** * * 哈斯拉链法 * author:huangkq1989 * blog: http://blog.csdn.net/kangquan2008 * *******************************/ #include <stdio.h> #include <stdlib.h> #define HASH_SIZE 11 #define TYPE int #define hash(data) \ (data)%HASH_SIZE typedef struct Node { TYPE data; struct Node * next; }Node; Node * node[HASH_SIZE]; void insert(TYPE data) { int index = hash(data); Node * new_node = malloc(sizeof(Node)); if(new_node == NULL) printf("memory alloc error"),exit(EXIT_FAILURE); new_node->data = data; new_node->next = NULL; if(node[index] != NULL) new_node->next = node[index]; node[index] = new_node; } Node* find(TYPE data) { int index = hash(data); Node * p = node[index]; while(p) { if(p->data == data) return p; p = p->next; } return NULL; } // 统计出现的次数 int count_appearanc_times(TYPE data) { int cnt = 0; int index = hash(data); Node * p = node[index]; while(p) { if(p->data == data) cnt ++; p = p->next; } return cnt; } int main() { TYPE data[] = {1,1,1,1,1,1,2,3123,4,2131,45,123,43543,6,1323,65, 66,7,6,76,8,127,48,237,623,96,2212,224,213,1224,234,1221,143,11,23,13111,311}; for(int i=0; i<sizeof(data)/sizeof(data[0]); i++) insert(data[i]); while(1) { printf("input your choice:1 for find,2 for count\n"); int choice; scanf("%d",&choice); switch(choice) { case 1: { printf("input the data you want to find:\n"); TYPE target; scanf("%d",&target); if(find(target)) printf("Got it: %d\n",target); else printf("Not found :(\n"); break; } case 2: { printf("input the data you want to count:\n"); TYPE target; scanf("%d",&target); int times; if(times = count_appearanc_times(target)) printf("Times of appearance: %d\n",times); else printf("Not found :(\n"); break; } default: break; } } return 0; }
