Number of Equivalent Domino Pairs

蹉跎岁月 岁月里蹉跎。今天补交一个题。

这个题在leetcode里是easy的难度,我确没有写完,真不愿意来承认啊。

题目如下:

Given a list of dominoesdominoes[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]

解:

一开始我就只想到了暴力double for loop,配合一个private method which is used to compare whether it is equal. 测试结果就是时间复杂度太高了,么有accept。

看了别人的解法,原来思路是这样的,as following code.

public int numEquivDominoPairs(int[][] dominoes) {
        int[] res = {0};
        Map<Integer, Integer> map =  new HashMap<>();
        int key = 0;
        for (int[] d : dominoes) {
            key = Math.min(d[0], d[1]) * 10 + Math.max(d[0], d[1]);
            map.put(key, map.getOrDefault(key, 0) + 1);
        }
        map.forEach((k, v) -> {
            res[0] += v * (v - 1) / 2;
        });
        return res[0];
    }

思路是把element array 转换成一个十进制数或者字符串。把得到的数或者字符串放在map里。然后用排列组合算出结果。res = C(2/v)

然后下面的解法就是直接累加计数得到结果:

    public int numEquivDominoPairs(int[][] dominoes) {
        int res = 0;
        int key = 0;
        Map<Integer, Integer> map = new HashMap<>();
        for (int[] d : dominoes) {
            key = Math.min(d[0], d[1]) * 10 + Math.max(d[0], d[1]);
            res += map.getOrDefault(key, 0);
            map.put(key, map.getOrDefault(key, 0) + 1);
        }
        return res;
    }

 

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