[2367] Number of Arithmetic Triplets

本來想說暴力下去解一定會超時，結果竟然沒有XD

但比較奇怪的是，C的solution 裡面大家都沒有用break！奇怪XD

但反正這是我的第一版 QQ 裡面的continue 看起來是可以拿掉就是了。（拿掉的版本就先不貼啦。）

int arithmeticTriplets(int* nums, int numsSize, int diff) {

int i=0;

int j=1;

int k=2;

int count =0;

for (i=0; i<numsSize-2;i++)

{

for (j=1;j< numsSize-1;j++)

{

if (nums[j]-nums[i]>diff)

break;

else if (nums[j]-nums[i] < diff)

continue;

else

{

for (int k=2;k<numsSize;k++)

{

if (nums[k]-nums[j]> diff)

break;

else if (nums[k]-nums[j] < diff)

continue;

else

{

count++;

break;

}

}

}

}

}

return count;

}

看了討論區，發現可以用hash table 寫！！！

那就變成.....

int arithmeticTriplets(int* nums, int numsSize, int diff) {

int i=0;

int count =0;

bool *hash = calloc (201, sizeof(bool));

for (int idx=0; idx<numsSize;idx++)

hash[nums[idx]] = true;

for (int i=0; i<199;i++)

{

if (hash[i] && i+diff+diff <=200)

{

if (hash[i+diff] && hash[i+diff+diff])

count++;

}

}

return count;

}

果然是easy題目，另人開心XD！！！