leetcode-1128-easy

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
Example 2:

Input: dominoes = [[1,2],[1,2],[1,1],[1,2],[2,2]]
Output: 3
Constraints:

1 <= dominoes.length <= 4 * 104
dominoes[i].length == 2
1 <= dominoes[i][j] <= 9

思路一：用 map 映射，原先想用 x * 10 + y 来作 map 的 key，但是 bug 比较多，最后直接用 string 一把梭了

    public static int numEquivDominoPairs(int[][] dominoes) {
        int ans = 0;

        Map<String, Integer> map = new HashMap<>();

        for (int[] dominoe : dominoes) {

            if (dominoe[0] > dominoe[1]) {
                map.compute(dominoe[0] + "," + dominoe[1], (k, v) -> v == null ? 1 : v + 1);
            } else {
                map.compute(dominoe[1] + "," + dominoe[0], (k, v) -> v == null ? 1 : v + 1);
            }
        }

        for (Map.Entry<String, Integer> e : map.entrySet()) {
            if (e.getValue() <= 1) {
                continue;
            }

            ans += e.getValue() * (e.getValue() - 1);
        }


        return ans / 2;
    }

思路二：补上官方的解

    public int numEquivDominoPairs(int[][] dominoes) {
        int[] num = new int[100];
        int ret = 0;
        for (int[] domino : dominoes) {
            int val = domino[0] < domino[1] ? domino[0] * 10 + domino[1] : domino[1] * 10 + domino[0];
            ret += num[val];
            num[val]++;
        }
        return ret;
    }