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NumberOfEquivalentDominoPairs.java
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import java.util.Arrays;
/**
* 1128. Number of Equivalent Domino Pairs
* 解法一:并查集。需要先排序,复杂度O(nlogn)。
* 解法二:哈希。时间复杂度 O(n).
* @author LBW
*/
public class NumberOfEquivalentDominoPairs {
// 解法一
public int numEquivDominoPairs(int[][] dominoes) {
int n = dominoes.length;
for (int i = 0; i < n; i++) {
if (dominoes[i][0] > dominoes[i][1]) {
int tmp = dominoes[i][0];
dominoes[i][0] = dominoes[i][1];
dominoes[i][1] = tmp;
}
}
Arrays.sort(dominoes, (o1, o2) -> (o1[0] * 10 + o1[1] - o2[0] * 10 - o2[1]));
DisjointSet disjointSet = new DisjointSet(n);
for (int i = 0; i < n - 1; i++) {
if (dominoes[i][0] == dominoes[i+1][0] && dominoes[i][1] == dominoes[i+1][1]) {
disjointSet.union(i, i+1);
}
}
return disjointSet.getResult();
}
static class DisjointSet {
int[] parent;
int[] size;
int n;
public DisjointSet(int n) {
this.n = n;
parent = new int[n];
size = new int[n];
for (int i = 0; i < n; i++) {
parent[i] = i;
size[i] = 1;
}
}
public int find(int k) {
if (parent[k] != k) {
parent[k] = find(parent[k]);
}
return parent[k];
}
public void union(int i, int j) {
int pi = find(i), pj = find(j);
if (pi != pj) {
parent[pj] = pi;
size[pi] += size[pj];
}
}
public int getResult() {
int res = 0;
for (int i = 0; i < n; i++) {
if (parent[i] == i) {
res += size[i] * (size[i] - 1) / 2;
}
}
return res;
}
}
// 解法二
public int numEquivDominoPairsTwo(int[][] dominoes) {
int[] hash = new int[100];
for (int i = 0; i < dominoes.length; i++) {
if (dominoes[i][0] > dominoes[i][1]) {
int tmp = dominoes[i][0];
dominoes[i][0] = dominoes[i][1];
dominoes[i][1] = tmp;
}
hash[dominoes[i][0] * 10 + dominoes[i][1]] += 1;
}
int res = 0;
for (int i = 0; i < 100; i++) {
if (hash[i] > 1) {
res += hash[i] * (hash[i] - 1) / 2;
}
}
return res;
}
public static void main(String[] args) {
NumberOfEquivalentDominoPairs numberOfEquivalentDominoPairs = new NumberOfEquivalentDominoPairs();
numberOfEquivalentDominoPairs.numEquivDominoPairs(new int[][]{{1, 2}, {1, 2},{1, 1}, {1, 2}, {2, 2}});
}
}