1128. The equivalent number of domino pair

Problem-solving ideas:

1 by means of the variable i, traversing the original array to dominoes [i] [0] and the dominoes [i] [1] as the smaller number of ten, as a larger number of bits

2. The new figures from the first step into a new array, while the current record number of new emerging digital

3. Since the digital array is greater than 0 and less than 10, the maximum value of the new number is 99.

4. from 0 to 99 traversing new array, if the number 1 is greater than the current number appears, the whole arrangement is calculated, the formula: n * (n-1) / 2

Time complexity of O (n)

Space complexity O (100)

Code:

class Solution {
    public int numEquivDominoPairs(int[][] dominoes) {
        int len=dominoes.length;
        int cnt=0;
        int k=0;
        int[] res=new int[100];
        for(int i=0;i<len;++i){
            int temp=0;
            //较小的数作为十位，较大的作为个位
            if(dominoes[i][0]<dominoes[i][1]){
                temp+=dominoes[i][0]*10+dominoes[i][1];
            }
            else{
                temp+=dominoes[i][1]*10+dominoes[i][0];
            }
            res[temp]++;
        }
        //找相同的数字的对数
        for(int i=0;i<100;++i){
            if(res[i]>1){
                cnt+=res[i]*(res[i]-1)/2;
            }
        }
        return cnt;
    }
}