题目;

法一 解题思路：

        |num[i] - num[j]| = k 可以理解为 num[j] = num[i] + k 和 num[j] = num[i] - k 两种情况。

int countKDifference(int* nums, int numsSize, int k) {int ans = 0;int hash[101];memset(hash, 0, sizeof(hash));for (int i = 0; i < numsSize; i++) {int x = nums[i] + k;if (x >= 1 && x <= 100) ans += hash[x];x = nums[i] - k;if (x >= 1 && x <= 100) ans += hash[x];hash[ nums[i] ]++;}return ans;
}

法二 动态分配

typedef struct  {int key;            int val;UT_hash_handle hh;
} HashEntry;int countKDifference(int* nums, int numsSize, int k){int ans = 0;HashEntry * cnt = NULL;for (int j = 0; j < numsSize; ++j) {HashEntry * pEntry = NULL;int curr = nums[j] - k;HASH_FIND(hh, cnt, &curr, sizeof(int),pEntry);if (NULL != pEntry) {ans += pEntry->val;}curr = nums[j] + k;HASH_FIND(hh, cnt, &curr, sizeof(int), pEntry);if (NULL != pEntry) {ans += pEntry->val;}HASH_FIND(hh, cnt, &nums[j], sizeof(int), pEntry);if (NULL == pEntry) {pEntry = (HashEntry *)malloc(sizeof(HashEntry));pEntry->key = nums[j];pEntry->val = 1;HASH_ADD(hh, cnt, key, sizeof(int), pEntry);} else {++pEntry->val;}}HashEntry * curr = NULL, * next = NULL;HASH_ITER(hh, cnt, curr, next) {HASH_DEL(cnt, curr);}return ans;
}