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| class Solution {
private int[] cityParent;
private int[] rank;
private void initialize() {
for (int i = 1; i < cityParent.length; i++) {
cityParent[i] = i;
rank[i] = 1;
}
}
private int find(int city) {
if (city == cityParent[city]) {
return city;
}
return cityParent[city] = find(cityParent[city]);
}
private void union(int city1, int city2) {
int city1Root = find(city1);
int city2Root = find(city2);
if (city1Root != city2Root) {
if (rank[city1Root] < rank[city2Root]) {
cityParent[city1Root] = city2Root;
} else if (rank[city1Root] > rank[city2Root]) {
cityParent[city2Root] = city1Root;
} else {
cityParent[city2Root] = city1Root;
rank[city1Root] += 1;
}
}
}
public int findCircleNum(int[][] isConnected) {
int numOfCities = isConnected.length;
cityParent = new int[numOfCities + 1];
rank = new int[numOfCities + 1];
initialize();
for (int i = 0; i < isConnected.length; i++) {
for (int j = 0; j < isConnected[0].length; j++) {
if (isConnected[i][j] == 1) {
union(i + 1, j + 1);
}
}
}
Set<Integer> rootCities = new HashSet<>();
for (int i = 1; i <= numOfCities; i++) {
rootCities.add(find(i));
}
return rootCities.size();
}
}
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Union Find. 我的首次尝试. 非常好用.
时间复杂度: O(n^2)
空间复杂度: O(n^2)
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| class Solution {
public int findCircleNum(int[][] isConnected) {
if (isConnected == null || isConnected.length == 0) {
return 0;
}
int n = isConnected.length;
UnionFind uf = new UnionFind(n);
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (isConnected[i][j] == 1) {
uf.union(i, j);
}
}
}
return uf.getCount();
}
class UnionFind {
private int[] root;
private int[] rank;
private int count;
UnionFind(int size) {
root = new int[size];
rank = new int[size];
count = size;
for (int i = 0; i < size; i++) {
root[i] = i;
rank[i] = 1;
}
}
int find(int x) {
if (x == root[x]) {
return x;
}
return root[x] = find(root[x]);
}
void union(int x, int y) {
int rootX = find(x);
int rootY = find(y);
if (rootX != rootY) {
if (rank[rootX] > rank[rootY]) {
root[rootY] = rootX;
} else if (rank[rootX] < rank[rootY]) {
root[rootX] = rootY;
} else {
root[rootY] = rootX;
rank[rootX] += 1;
}
count--;
}
}
int getCount() {
return count;
}
}
}
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这是官方的题解. 用一个inner class来表示union find. 这个class里面还记录了一个count变量, 它一开始被初始化为city的个数. 当出现成功的union后, 我们让count减1. 这个做法学到了.
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| class Solution {
private static class UnionFind {
private int[] cityParent;
private int[] rank;
private int count;
private UnionFind(int size) {
cityParent = new int[size + 1];
rank = new int[size + 1];
for (int i = 1; i < cityParent.length; i++) {
cityParent[i] = i;
rank[i] = 1;
}
count = size;
}
private int find(int city) {
if (city == cityParent[city]) {
return city;
}
return cityParent[city] = find(cityParent[city]);
}
private void union(int city1, int city2) {
int city1Root = find(city1);
int city2Root = find(city2);
if (city1Root != city2Root) {
if (rank[city1Root] < rank[city2Root]) {
cityParent[city1Root] = city2Root;
} else if (rank[city1Root] > rank[city2Root]) {
cityParent[city2Root] = city1Root;
} else {
cityParent[city2Root] = city1Root;
rank[city1Root] += 1;
}
count -= 1;
}
}
private int getCount() {
return count;
}
}
public int findCircleNum(int[][] isConnected) {
int numOfCities = isConnected.length;
UnionFind uf = new UnionFind(numOfCities);
for (int i = 0; i < isConnected.length; i++) {
for (int j = 0; j < isConnected[0].length; j++) {
if (isConnected[i][j] == 1) {
uf.union(i + 1, j + 1);
}
}
}
return uf.getCount();
}
}
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这个是我自己又把union find打包成class重写了一遍. 这就是模板.
