题目:
题解:
cpp
struct hash_table {
int key;
int val;
// 查看 https://troydhanson.github.io/uthash/ 了解更多
UT_hash_handle hh;
};
typedef struct hash_table* hash_ptr;
struct pair {
int first;
int second;
};
void swap(struct pair* a, struct pair* b) {
struct pair t = *a;
*a = *b, *b = t;
}
void sort(struct pair* v, int start, int end, int* ret, int* retSize, int k) {
int picked = rand() % (end - start + 1) + start;
swap(&v[picked], &v[start]);
int pivot = v[start].second;
int index = start;
for (int i = start + 1; i <= end; i++) {
// 使用双指针把不小于基准值的元素放到左边,
// 小于基准值的元素放到右边
if (v[i].second >= pivot) {
swap(&v[index + 1], &v[i]);
index++;
}
}
swap(&v[start], &v[index]);
if (k <= index - start) {
// 前 k 大的值在左侧的子数组里
sort(v, start, index - 1, ret, retSize, k);
} else {
// 前 k 大的值等于左侧的子数组全部元素
// 加上右侧子数组中前 k - (index - start + 1) 大的值
for (int i = start; i <= index; i++) {
ret[(*retSize)++] = v[i].first;
}
if (k > index - start + 1) {
sort(v, index + 1, end, ret, retSize, k - (index - start + 1));
}
}
}
int* topKFrequent(int* nums, int numsSize, int k, int* returnSize) {
hash_ptr head = NULL;
hash_ptr p = NULL, tmp = NULL;
// 获取每个数字出现次数
for (int i = 0; i < numsSize; i++) {
HASH_FIND_INT(head, &nums[i], p);
if (p == NULL) {
p = malloc(sizeof(struct hash_table));
p->key = nums[i];
p->val = 1;
HASH_ADD_INT(head, key, p);
} else {
p->val++;
}
}
struct pair values[numsSize];
int valuesSize = 0;
HASH_ITER(hh, head, p, tmp) {
values[valuesSize].first = p->key;
values[valuesSize++].second = p->val;
}
int* ret = malloc(sizeof(int) * k);
*returnSize = 0;
sort(values, 0, valuesSize - 1, ret, returnSize, k);
return ret;
}