题目：

法一 解题思路：

        由于 nums.length 小于100，新建数组 num[101]，用来遍历存放 nums[i]出现的次数。

int sumOfUnique(int* nums, int numsSize) {int result = 0;int num[101] = {0};  // memset(num, 0, sizof(num));for (int i = 0; i < numsSize; i++) {num[nums[i]]++;}for (int i = 0; i < 101; i++) {if (num[i] == 1) result += i;}return result;
}

法二 哈希表（官方）解题思路：

typedef struct {int key;int val;UT_hash_handle hh;
}hashEntry;int sumOfUnique(int* nums, int numsSize) {hashEntry * cnt = NULL;int ans = 0;for (int i = 0; i < numsSize; ++i) {hashEntry * qEntry = NULL;HASH_FIND(hh, cnt, &nums[i], sizeof(int), qEntry);if (qEntry == NULL) {hashEntry * qEntry = (hashEntry*) malloc(sizeof(hashEntry));qEntry -> key = nums[i];qEntry -> val = 1;ans += nums[i];HASH_ADD(hh, cnt, key,sizeof(int), qEntry);}else if(qEntry->val == 1){ans -= nums[i];qEntry -> val = 2;}}hashEntry * curr = NULL, * next = NULL;HASH_ITER(hh, cnt, curr, next){HASH_DEL(cnt, curr);free(curr);}return ans;
}