Given two arrays, write a function to compute their intersection.
Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2].
Note:
Each element in the result should appear as many times as it shows in both arrays.
The result can be in any order.
Follow up:
What if the given array is already sorted? How would you optimize your algorithm?
What if nums1's size is small compared to nums2's size? Which algorithm is better?
What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?
#define HASH_SIZE 1000
struct hashnode{
int val;
int times;
struct hashnode *next;
};
struct hash{
struct hashnode * hash_table;
int (*hash_func)(int);
};
int hash_func(int val)
{
return abs(val)%HASH_SIZE;
}
void hash_add(struct hash * h, int val)
{
int index = h->hash_func(val);
struct hashnode *dummy = &(h->hash_table[index]);
if(dummy->next == NULL){
struct hashnode * node = calloc(1, sizeof(struct hashnode));
node->val = val;
node->times = 1;
dummy->next = node;
}else{
struct hashnode *p = dummy->next;
while(p){
if(p->val == val){
p->times += 1;
return;
}
p = p->next;
}
struct hashnode * node = calloc(1, sizeof(struct hashnode));
node->val = val;
node->times = 1;
node->next = dummy->next;
dummy->next = node;
}
}
struct hash * hash_init()
{
struct hash * h = calloc(1, sizeof(struct hash));
h->hash_table = calloc(HASH_SIZE, sizeof(struct hashnode));
h->hash_func = hash_func;
return h;
}
/**
* Return an array of size *returnSize.
* Note: The returned array must be malloced, assume caller calls free().
*/
int* intersect(int* nums1, int nums1Size, int* nums2, int nums2Size, int* returnSize) {
struct hash * h = hash_init();
int * ret = calloc(0, sizeof(int));
*returnSize = 0;
for(int i = 0; i < nums1Size; i++){
hash_add(h, nums1[i]);
}
for(int i = 0; i < nums2Size; i++){
int index = h->hash_func(nums2[i]);
struct hashnode *dummy = &(h->hash_table[index]);
if(dummy->next == NULL){
}else{
struct hashnode *p = dummy->next;
struct hashnode *prev = dummy;
while(p){
if(p->val == nums2[i]){
//save val. intersection node found
*returnSize += 1;
ret = realloc(ret, (*returnSize)*(sizeof(int)));
ret[*returnSize - 1] = nums2[i];
p->times -= 1;
if(0 == p->times){
//remove p
prev->next = p->next;
free(p);
}
}
p = p->next;
prev = prev->next;
}
}
}
return ret;
}
350. Intersection of Two Arrays II
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