feat: add solutions to lcci problem: No.16.19.Pond Sizes

Author: yanglbme

lcci/16.19.Pond Sizes/README.md

@@ -26,22 +26,314 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+并查集。
+
+并查集模板：
+
+模板 1——朴素并查集：
+
+```python
+# 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：
+
+```python
+# 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：
+
+```python
+# 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)] = distance
+```
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
+class 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 res
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class 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;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+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;
+    }
+};
+```
+
+### **Go**
+
+```go
+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
+}
 ```
 
 ### **...**