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].
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Example 1:
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Input: dominoes = [[1,2],[2,1],[3,4],[5,6]]
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Output: 1
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Constraints:
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1 <= dominoes.length <= 40000
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1 <= dominoes[i][j] <= 9
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可以先按大小排个序，然后建立一个map，统计次数大于1的元素个数，每个元素的对数为C(2,n)
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class Solution {
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public:
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int numEquivDominoPairs(vector<vector<int>>& dominoes) {
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int res=0;
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if(dominoes.size()<2) return res;
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for(auto &v: dominoes)
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if(v[0]>v[1]) swap(v[0],v[1]);
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map<vector<int>, int> domin_map;
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for(auto v:dominoes)
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domin_map[v]++;
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for(auto v:domin_map)
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if(v.second >1) res+= v.second * (v.second -1);
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return res/2;
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}
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};
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执行用时 :68 ms, 在所有 C++ 提交中击败了76.11%的用户
内存消耗 :23.5 MB, 在所有 C++ 提交中击败了100.00%的用户