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
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; } }