你有一个用于表示一片土地的整数矩阵land，该矩阵中每个点的值代表对应地点的海拔高度。若值为0则表示水域。由垂直、水平或对角连接的水域为池塘。池塘的大小是指相连接的水域的个数。编写一个方法来计算矩阵中所有池塘的大小，返回值需要从小到大排序。
示例：
输入： [ [0,2,1,0], [0,1,0,1], [1,1,0,1], [0,1,0,1] ] 输出： [1,2,4]
提示：
0 < len(land) <= 10000 < len(land[i]) <= 1000
并查集。
并查集模板：
模板 1——朴素并查集：
# 初始化，p存储每个点的父节点
p = list(range(n))
# 返回x的祖宗节点
def find(x):
if p[x] != x:
# 路径压缩
p[x] = find(p[x])
return p[x]
# 合并a和b所在的两个集合
p[find(a)] = find(b)模板 2——维护 size 的并查集：
# 初始化，p存储每个点的父节点，size只有当节点是祖宗节点时才有意义，表示祖宗节点所在集合中，点的数量
p = list(range(n))
size = [1] * n
# 返回x的祖宗节点
def find(x):
if p[x] != x:
# 路径压缩
p[x] = find(p[x])
return p[x]
# 合并a和b所在的两个集合
if find(a) != find(b):
size[find(b)] += size[find(a)]
p[find(a)] = find(b)模板 3——维护到祖宗节点距离的并查集：
# 初始化，p存储每个点的父节点，d[x]存储x到p[x]的距离
p = list(range(n))
d = [0] * n
# 返回x的祖宗节点
def find(x):
if p[x] != x:
t = find(p[x])
d[x] += d[p[x]]
p[x] = t
return p[x]
# 合并a和b所在的两个集合
p[find(a)] = find(b)
d[find(a)] = distanceclass Solution:
def pondSizes(self, land: List[List[int]]) -> List[int]:
m, n = len(land), len(land[0])
p = list(range(m * n))
size = [1] * (m * n)
def find(x):
if p[x] != x:
p[x] = find(p[x])
return p[x]
def union(a, b):
pa, pb = find(a), find(b)
if pa == pb:
return
size[pb] += size[pa]
p[pa] = pb
for i in range(m):
for j in range(n):
if land[i][j] != 0:
continue
idx = i * n + j
if i < m - 1 and land[i + 1][j] == 0:
union(idx, (i + 1) * n + j)
if j < n - 1 and land[i][j + 1] == 0:
union(idx, i * n + j + 1)
if i < m - 1 and j < n - 1 and land[i + 1][j + 1] == 0:
union(idx, (i + 1) * n + j + 1)
if i < m - 1 and j > 0 and land[i + 1][j - 1] == 0:
union(idx, (i + 1) * n + j - 1)
s = set()
res = []
for i in range(m * n):
if land[i // n][i % n] != 0:
continue
root = find(i)
if root not in s:
s.add(root)
res.append(size[root])
res.sort()
return resclass Solution {
private int[] p;
private int[] size;
public int[] pondSizes(int[][] land) {
int m = land.length, n = land[0].length;
p = new int[m * n];
size = new int[m * n];
for (int i = 0; i < p.length; ++i) {
p[i] = i;
size[i] = 1;
}
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (land[i][j] != 0) {
continue;
}
int idx = i * n + j;
if (i < m - 1 && land[i + 1][j] == 0) {
union(idx, (i + 1) * n + j);
}
if (j < n - 1 && land[i][j + 1] == 0) {
union(idx, i * n + j + 1);
}
if (i < m - 1 && j < n - 1 && land[i + 1][j + 1] == 0) {
union(idx, (i + 1) * n + j + 1);
}
if (i < m - 1 && j > 0 && land[i + 1][j - 1] == 0) {
union(idx, (i + 1) * n + j - 1);
}
}
}
Set<Integer> s = new HashSet<>();
List<Integer> t = new ArrayList<>();
for (int i = 0; i < m * n; ++i) {
if (land[i / n][i % n] != 0) {
continue;
}
int root = find(i);
if (!s.contains(root)) {
s.add(root);
t.add(size[root]);
}
}
Collections.sort(t);
int[] res = new int[t.size()];
for (int i = 0; i < res.length; ++i) {
res[i] = t.get(i);
}
return res;
}
private int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}
private void union(int a, int b) {
int pa = find(a), pb = find(b);
if (pa == pb) {
return;
}
size[pb] += size[pa];
p[pa] = pb;
}
}class Solution {
public:
vector<int> p;
vector<int> size;
vector<int> pondSizes(vector<vector<int>>& land) {
int m = land.size(), n = land[0].size();
for (int i = 0; i < m * n; ++i) {
p.push_back(i);
size.push_back(1);
}
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (land[i][j] != 0) continue;
int idx = i * n + j;
if (i < m - 1 && land[i + 1][j] == 0) unite(idx, (i + 1) * n + j);
if (j < n - 1 && land[i][j + 1] == 0) unite(idx, i * n + j + 1);
if (i < m - 1 && j < n - 1 && land[i + 1][j + 1] == 0) unite(idx, (i + 1) * n + j + 1);
if (i < m - 1 && j > 0 && land[i + 1][j - 1] == 0) unite(idx, (i + 1) * n + j - 1);
}
}
unordered_set<int> s;
vector<int> res;
for (int i = 0; i < m * n; ++i) {
if (land[i / n][i % n] != 0) continue;
int root = find(i);
if (s.find(root) == s.end()) {
s.insert(root);
res.push_back(size[root]);
}
}
sort(res.begin(), res.end());
return res;
}
int find(int x) {
if (p[x] != x) p[x] = find(p[x]);
return p[x];
}
void unite(int a, int b) {
int pa = find(a), pb = find(b);
if (pa == pb) return;
size[pb] += size[pa];
p[pa] = pb;
}
};var p []int
var size []int
func pondSizes(land [][]int) []int {
m, n := len(land), len(land[0])
p = make([]int, m*n)
size = make([]int, m*n)
for i := 0; i < m*n; i++ {
p[i] = i
size[i] = 1
}
for i := 0; i < m; i++ {
for j := 0; j < n; j++ {
if land[i][j] != 0 {
continue
}
idx := i*n + j
if i < m-1 && land[i+1][j] == 0 {
union(idx, (i+1)*n+j)
}
if j < n-1 && land[i][j+1] == 0 {
union(idx, i*n+j+1)
}
if i < m-1 && j < n-1 && land[i+1][j+1] == 0 {
union(idx, (i+1)*n+j+1)
}
if i < m-1 && j > 0 && land[i+1][j-1] == 0 {
union(idx, (i+1)*n+j-1)
}
}
}
s := make(map[int]bool)
var res []int
for i := 0; i < m*n; i++ {
if land[i/n][i%n] != 0 {
continue
}
root := find(i)
if !s[root] {
s[root] = true
res = append(res, size[root])
}
}
sort.Ints(res)
return res
}
func find(x int) int {
if p[x] != x {
p[x] = find(p[x])
}
return p[x]
}
func union(a, b int) {
pa, pb := find(a), find(b)
if pa == pb {
return
}
size[pb] += size[pa]
p[pa] = pb
}