2364. Count Number of Bad Pairs
## 思路
Bad pair: i < j and j - i != nums[j] - nums[i]
移项后, nums[j] - j != nums[i] - i
所以建个counter, 纪录(nums[i]-i) 的个数
当前index i的 bad pair数就会是i - counter[nums[i]-i]
## Code
```cpp
class Solution {
public:
long long countBadPairs(vector<int>& nums) {
// nums[j] - j != nums[i] - i
unordered_map<int, int> counter;
long long res = 0;
for (int i=0; i<nums.size(); ++i) {
res += (i - counter[nums[i]-i]++);
}
return res;
}
};
```