feat: add solutions to lc problem: No.0778.Swim in Rising Water

Author: yanglbme

solution/0700-0799/0778.Swim in Rising Water/README.md

@@ -51,19 +51,105 @@
 	<li><code>grid[i][j]</code> 是 <code>[0, ..., N*N - 1]</code> 的排列。</li>
 </ol>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
 
+并查集。
+
+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 swimInWater(self, grid: List[List[int]]) -> int:
+        n = len(grid)
+        p = list(range(n * n))
+
+        def find(x):
+            if p[x] != x:
+                p[x] = find(p[x])
+            return p[x]
+
+        def index(i, j):
+            return i * n + j
+
+        def check(i, j):
+            return 0 <= i < n and 0 <= j < n
+
+        hi = [0] * (n * n)
+        for i in range(n):
+            for j in range(n):
+                hi[grid[i][j]] = index(i, j)
+        for h in range(n * n):
+            x, y = hi[h] // n, hi[h] % n
+            for a, b in [(0, -1), (0, 1), (1, 0), (-1, 0)]:
+                x1, y1 = x + a, y + b
+                if check(x1, y1) and grid[x1][y1] <= h:
+                    p[find(index(x1, y1))] = find(hi[h])
+                if find(0) == find(n * n - 1):
+                    return h
+        return -1
 ```
 
 ### **Java**
@@ -72,54 +158,149 @@
 
 ```java
 class Solution {
-    // x、y方向向量
-    public static final int[] dx = {0, 0, 1, -1};
-    public static final int[] dy = {1, -1, 0, 0};
-    /**
-     * https://blog.csdn.net/fuxuemingzhu/article/details/82926674
-     * <p>
-     * 参考这篇文章的第二种解题方法做的
-     * <p>
-     * 通过优先级队列找寻局部最优解  最终的得到的结果就是全局最优解
-     *
-     * @param grid
-     * @return
-     */
-    // 以grid左上角为原点，横向为X轴，纵向为Y轴
+    private int[] p;
+    private int n;
+    private int[][] dirs = new int[][]{{0, -1}, {0, 1}, {1, 0}, {-1, 0}};
+
     public int swimInWater(int[][] grid) {
-        // 定义一个优先级队列  按照h从小到大排列
-        Queue<Pair<Integer, Pair<Integer, Integer>>> queue = new PriorityQueue<>(Comparator.comparing(Pair::getKey));
-        queue.add(new Pair<>(grid[0][0], new Pair<>(0, 0)));
-        // 已经遍历过的点
-        Set<Pair<Integer, Integer>> visitSet = new HashSet<>();
-        visitSet.add(new Pair<>(0, 0));
-
-        int res = 0;
-        int length = grid.length;
-
-        while (!queue.isEmpty()) {
-            Pair<Integer, Pair<Integer, Integer>> top = queue.poll();
-            Integer x = top.getValue().getKey();
-            Integer y = top.getValue().getValue();
-            res = Math.max(res, top.getKey());
-            // 2 <= N <= 50 这个范围内可以直接使用==进行Integer的比较
-            if (x == top.getValue().getValue() && y == length - 1) {
-                break;
+        n = grid.length;
+        p = new int[n * n];
+        for (int i = 0; i < p.length; ++i) {
+            p[i] = i;
+        }
+        int[] hi = new int[n * n];
+        for (int i = 0; i < n; ++i) {
+            for (int j = 0; j < n; ++j) {
+                hi[grid[i][j]] = index(i, j);
             }
-
-            for (int i = 0; i < 4; i++) {
-                int newY = y + dy[i];
-                int newX = x + dx[i];
-                if (newX < 0 || newY < 0 || newX >= length || newY >= length || visitSet.contains(new Pair<>(newX, newY))) {
-                    // 直接忽略
-                    continue;
+        }
+        for (int h = 0; h < n * n; ++h) {
+            int x = hi[h] / n, y = hi[h] % n;
+            for (int[] dir : dirs) {
+                int x1 = x + dir[0], y1 = y + dir[1];
+                if (check(x1, y1) && grid[x1][y1] <= h) {
+                    p[find(index(x1, y1))] = find(hi[h]);
+                }
+                if (find(0) == find(n * n - 1)) {
+                    return h;
                 }
-                queue.add(new Pair<>(grid[newX][newY], new Pair<>(newX, newY)));
-                visitSet.add(new Pair<>(newX, newY));
             }
         }
-        return res;
+        return -1;
+    }
+
+    private int find(int x) {
+        if (p[x] != x) {
+            p[x] = find(p[x]);
+        }
+        return p[x];
+    }
+
+    private int index(int i, int j) {
+        return i * n + j;
     }
+
+    private boolean check(int i, int j) {
+        return i >= 0 && i < n && j >= 0 && j < n;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    vector<int> p;
+    int n;
+    int dirs[4][2] = {{0, -1}, {0, 1}, {1, 0}, {-1, 0}};
+
+    int swimInWater(vector<vector<int>> &grid) {
+        n = grid.size();
+        for (int i = 0; i < n * n; ++i)
+            p.push_back(i);
+        vector<int> hi(n * n, 0);
+        for (int i = 0; i < n; ++i)
+            for (int j = 0; j < n; ++j)
+                hi[grid[i][j]] = index(i, j);
+        for (int h = 0; h < n * n; ++h)
+        {
+            int x = hi[h] / n, y = hi[h] % n;
+            for (auto dir : dirs)
+            {
+                int x1 = x + dir[0], y1 = y + dir[1];
+                if (check(x1, y1) && grid[x1][y1] <= h)
+                    p[find(index(x1, y1))] = find(hi[h]);
+                if (find(0) == find(n * n - 1))
+                    return h;
+            }
+        }
+        return -1;
+    }
+
+    int find(int x) {
+        if (p[x] != x)
+            p[x] = find(p[x]);
+        return p[x];
+    }
+
+    int index(int i, int j) {
+        return i * n + j;
+    }
+
+    bool check(int i, int j) {
+        return i >= 0 && i < n && j >= 0 && j < n;
+    }
+};
+```
+
+### **Go**
+
+```go
+var p []int
+var n int
+
+func swimInWater(grid [][]int) int {
+	n = len(grid)
+	p = make([]int, n*n)
+	hi := make([]int, n*n)
+	for i := 0; i < len(p); i++ {
+		p[i] = i
+	}
+	for i := 0; i < n; i++ {
+		for j := 0; j < n; j++ {
+			hi[grid[i][j]] = index(i, j)
+		}
+	}
+	dirs := [4][2]int{{0, -1}, {0, 1}, {1, 0}, {-1, 0}}
+	for h := 0; h < n*n; h++ {
+		x, y := hi[h]/n, hi[h]%n
+		for _, dir := range dirs {
+			x1, y1 := x+dir[0], y+dir[1]
+			if check(x1, y1) && grid[x1][y1] <= h {
+				p[find(index(x1, y1))] = find(hi[h])
+			}
+			if find(0) == find(n*n-1) {
+				return h
+			}
+		}
+	}
+	return -1
+}
+
+func find(x int) int {
+	if p[x] != x {
+		p[x] = find(p[x])
+	}
+	return p[x]
+}
+
+func index(i, j int) int {
+	return i*n + j
+}
+
+func check(i, j int) bool {
+	return i >= 0 && i < n && j >= 0 && j < n
 }
 ```