题目：

题解：

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;
}