1128. Number of Equivalent Domino Pairs

Given a list of dominoes, dominoes[i] = [a, b] is equivalent to dominoes[j] = [c, d] if and only if either (a==c and b==d), or (a==d and b==c) - that is, one domino can be rotated to be equal to another domino.

Return the number of pairs (i, j) for which 0 <= i < j < dominoes.length, and dominoes[i] is equivalent to dominoes[j].

 

Example 1:

Input: dominoes = [[1,2],[2,1],[3,4],[5,6]]
Output: 1
 

Constraints:

1 <= dominoes.length <= 40000
1 <= dominoes[i][j] <= 9

这题类似于排序，只是坐标之间反的也是相等的。那么我们按照这个规则进行排序后计算一下重复的个数即可。

class NumberOfEquivalentDominoPairs : public Solution {
public:
    void Exec() {
        vector<vector<int>> input;
        input.push_back(vector<int>{1,2});
        input.push_back(vector<int>{1,2});
        input.push_back(vector<int>{1,1});
        input.push_back(vector<int>{1,2});
        input.push_back(vector<int>{2,2});
        auto res = numEquivDominoPairs(input);
    }
    int numEquivDominoPairs(vector<vector<int>>& dominoes) {
        int res = 0;

        std::sort(dominoes.begin(), dominoes.end(), [](vector<int> &a, vector<int> &b) -> bool {
            if (a[0] > a[1]) {
                std::swap(a[0], a[1]);
            }
            if (b[0] > b[1]) {
                std::swap(b[0], b[1]);
            }
            if (a[0] == b[0]) {
                return a[1] < b[1];
            }
            return a[0] < b[0];
        });

        int conts = 0;
        for (int i = 1; i <= dominoes.size(); i++) {
            if (i < dominoes.size() && dominoes[i][0] == dominoes[i - 1][0] &&
                dominoes[i][1] == dominoes[i - 1][1]) {
                conts++;
            } else {
                if (conts != 0) {
                    res += conts * (1 + conts) / 2;
                }
                conts = 0;
            }
        }
        return res;
    }
};