标签：pair integer opened == either etc lap ota alt
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==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 <= 400001 <= 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; } }
标签：pair integer opened == either etc lap ota alt
原文地址：https://www.cnblogs.com/Asimple/p/11258542.html
