feat: add solutions to lc problem: No.0827.Making A Large Island

Author: yanglbme

solution/0100-0199/0128.Longest Consecutive Sequence/README.md

@@ -94,11 +94,9 @@ class Solution:
         s, res = set(nums), 0
         for num in nums:
             if num - 1 not in s:
-                t = 1
-                next = num + 1
+                t, next = 1, num + 1
                 while next in s:
-                    t += 1
-                    next += 1
+                    t, next = t + 1, next + 1
                 res = max(res, t)
         return res
 ```

solution/0100-0199/0128.Longest Consecutive Sequence/README_EN.md

@@ -72,11 +72,9 @@ class Solution:
         s, res = set(nums), 0
         for num in nums:
             if num - 1 not in s:
-                t = 1
-                next = num + 1
+                t, next = 1, num + 1
                 while next in s:
-                    t += 1
-                    next += 1
+                    t, next = t + 1, next + 1
                 res = max(res, t)
         return res
 ```

solution/0100-0199/0128.Longest Consecutive Sequence/Solution.py

@@ -3,10 +3,8 @@ def longestConsecutive(self, nums: List[int]) -> int:
         s, res = set(nums), 0
         for num in nums:
             if num - 1 not in s:
-                t = 1
-                next = num + 1
+                t, next = 1, num + 1
                 while next in s:
-                    t += 1
-                    next += 1
+                    t, next = t + 1, next + 1
                 res = max(res, t)
         return res

solution/0800-0899/0827.Making A Large Island/README.md

@@ -39,27 +39,357 @@
 	<li><code>0 &lt;= grid[i][j] &lt;= 1</code></li>
 </ul>
 
-
 ## 解法
 
 <!-- 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 largestIsland(self, grid: List[List[int]]) -> int:
+        n = len(grid)
+        p = list(range(n * n))
+        size = [1] * (n * 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:
+                size[pb] += size[pa]
+                p[pa] = pb
+
+        def check(i, j):
+            return 0 <= i < n and 0 <= j < n and grid[i][j] == 1
+
+        for i in range(n):
+            for j in range(n):
+                if grid[i][j] == 1:
+                    for x, y in [[1, 0], [0, 1]]:
+                        if check(i + x, j +y):
+                            union(i * n + j, (i + x) * n + j + y)
+
+        res = max(size)
+        for i in range(n):
+            for j in range(n):
+                if grid[i][j] == 0:
+                    t = 1
+                    s = set()
+                    for x, y in [[0, 1], [0, -1], [1, 0], [-1, 0]]:
+                        if check(i + x, j + y):
+                            root = find((i + x) * n + j + y)
+                            if root not in s:
+                                t += size[root]
+                                s.add(root)
+                    res = max(res, t)
+        return res
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    private int n;
+    private int[] p;
+    private int[] size;
+    private int mx;
+    private int[][] grid;
+    private int[][] dirs = new int[][]{{0, -1}, {0, 1}, {1, 0}, {-1, 0}};
+
+    public int largestIsland(int[][] grid) {
+        n = grid.length;
+        mx = 1;
+        this.grid = grid;
+        p = new int[n * n];
+        size = new int[n * n];
+        for (int i = 0; i < p.length; ++i) {
+            p[i] = i;
+            size[i] = 1;
+        }
+        for (int i = 0; i < n; ++i) {
+            for (int j = 0; j < n; ++j) {
+                if (grid[i][j] == 1) {
+                    for (int[] e : dirs) {
+                        if (check(i + e[0], j + e[1])) {
+                            union(i * n + j, (i + e[0]) * n + j + e[1]);
+                        }
+                    }
+                }
+            }
+        }
+        int res = mx;
+        for (int i = 0; i < n; ++i) {
+            for (int j = 0; j < n; ++j) {
+                if (grid[i][j] == 0) {
+                    int t = 1;
+                    Set<Integer> s = new HashSet<>();
+                    for (int[] e : dirs) {
+                        if (check(i + e[0], j + e[1])) {
+                            int root = find((i + e[0]) * n + j + e[1]);
+                            if (!s.contains(root)) {
+                                t += size[root];
+                                s.add(root);
+                            }
+                        }
+                    }
+                    res = Math.max(res, t);
+                }
+            }
+        }
+        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) {
+            size[pb] += size[pa];
+            mx = Math.max(mx, size[pb]);
+            p[pa] = pb;
+        }
+    }
+
+    private boolean check(int i, int j) {
+        return i >= 0 && i < n && j >= 0 && j < n && grid[i][j] == 1;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    vector<int> p;
+    vector<int> size;
+    int n, mx;
+    int dirs[4][2] = {{0, -1}, {0, 1}, {1, 0}, {-1, 0}};
+
+    int largestIsland(vector<vector<int>>& grid) {
+        n = grid.size();
+        mx = 1;
+        p.resize(n * n);
+        size.resize(n * n);
+        for (int i = 0; i < p.size(); ++i)
+        {
+            p[i] = i;
+            size[i] = 1;
+        }
+        for (int i = 0; i < n; ++i)
+        {
+            for (int j = 0; j < n; ++j)
+            {
+                if (grid[i][j] == 1)
+                {
+                    for (auto e : dirs)
+                    {
+                        if (check(i + e[0], j + e[1], grid)) unite(i * n + j, (i + e[0]) * n + j + e[1]);
+                    }
+                }
+            }
+        }
+        int res = mx;
+        for (int i = 0; i < n; ++i)
+        {
+            for (int j = 0; j < n; ++j)
+            {
+                if (grid[i][j] == 0)
+                {
+                    int t = 1;
+                    unordered_set<int> s;
+                    for (auto e : dirs)
+                    {
+                        if (check(i + e[0], j + e[1], grid))
+                        {
+                            int root = find((i + e[0]) * n + j + e[1]);
+                            if (!s.count(root)) {
+                                t += size[root];
+                                s.insert(root);
+                            }
+                        }
+                    }
+                    res = max(res, t);
+                }
+            }
+        }
+        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)
+        {
+            size[pb] += size[pa];
+            mx = max(mx, size[pb]);
+            p[pa] = pb;
+        }
+    }
+
+    bool check(int i, int j, vector<vector<int>>& grid) {
+        return i >= 0 && i < n && j >= 0 && j < n && grid[i][j] == 1;
+    }
+};
+```
+
+### **Go**
+
+```go
+var p []int
+var size []int
+var n int
+var mx int
+
+func largestIsland(grid [][]int) int {
+	n, mx = len(grid), 1
+	p = make([]int, n*n)
+	size = make([]int, n*n)
+	for i := 0; i < len(p); i++ {
+		p[i] = i
+		size[i] = 1
+	}
+
+	dirs := [4][2]int{{0, -1}, {0, 1}, {1, 0}, {-1, 0}}
+	for i := 0; i < n; i++ {
+		for j := 0; j < n; j++ {
+			if grid[i][j] == 1 {
+				for _, e := range dirs {
+					if check(i+e[0], j+e[1], grid) {
+						union(i*n+j, (i+e[0])*n+j+e[1])
+					}
+				}
+			}
+		}
+	}
+	res := mx
+	for i := 0; i < n; i++ {
+		for j := 0; j < n; j++ {
+			if grid[i][j] == 0 {
+				t := 1
+				s := make(map[int]bool)
+				for _, e := range dirs {
+					if check(i+e[0], j+e[1], grid) {
+						root := find((i+e[0])*n + j + e[1])
+						if !s[root] {
+							t += size[root]
+							s[root] = true
+						}
+					}
+				}
+				res = max(res, t)
+			}
+		}
+	}
+	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 {
+		size[pb] += size[pa]
+		mx = max(mx, size[pb])
+		p[pa] = pb
+	}
+}
+
+func check(i, j int, grid [][]int) bool {
+	return i >= 0 && i < n && j >= 0 && j < n && grid[i][j] == 1
+}
 
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
 ```
 
 ### **...**