LeetCode Weekly Contest 146

1128. Number of Equivalent Domino Pairs

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==dand 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

题目大意:给你一个数字键值对数组,让你判断有多少个数组元素是相等的。两个键值对相等的条件是:键1=键2并且值1=值2或者键1=值2并且键2=值1。

思路:将键值的和中放前面加上“_”以及键值中较小的数作为map的key存入map,然后遍历map求和就好。(其实可以将键值中较小的数乘100或者1000然后加上大的那个数作为key,但是竞赛的时候,脑子没转过来,就用了这个方法,也是AC了)

class Solution {
    public int numEquivDominoPairs(int[][] dominoes) {
        Map<String, Integer> map = new HashMap<>();
        int len = dominoes.length;
        for(int i=0; i<len; i++) {
            int a = dominoes[i][0];
            int b = dominoes[i][1];
            String te = a+b + "-";
            if( a > b ) te = te + b;
            else te = te + a;
            Integer cnt = map.get(te);
            if( cnt==null ) {
                map.put(te, 1);
            } else {
                map.put(te, cnt+1);
            }
        }
        int sum = 0;
        for(Map.Entry<String, Integer> entry : map.entrySet() ) {
            int v = entry.getValue();
            sum += (v*(v-1))/2;
        }
        return sum;
    }
}
View Code

 

posted @ 2019-07-28 12:49  Asimple  阅读(130)  评论(0编辑  收藏  举报