diff --git a/solution/0700-0799/0778.Swim in Rising Water/README.md b/solution/0700-0799/0778.Swim in Rising Water/README.md
index 8b366b4866899..2168fe1d291e0 100644
--- a/solution/0700-0799/0778.Swim in Rising Water/README.md
+++ b/solution/0700-0799/0778.Swim in Rising Water/README.md
@@ -64,7 +64,56 @@
```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
+ *
+ * 参考这篇文章的第二种解题方法做的
+ *
+ * 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
+ public int swimInWater(int[][] grid) {
+ // 定义一个优先级队列 按照h从小到大排列
+ Queue>> 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> visitSet = new HashSet<>();
+ visitSet.add(new Pair<>(0, 0));
+
+ int res = 0;
+ int length = grid.length;
+
+ while (!queue.isEmpty()) {
+ Pair> 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;
+ }
+
+ 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;
+ }
+ queue.add(new Pair<>(grid[newX][newY], new Pair<>(newX, newY)));
+ visitSet.add(new Pair<>(newX, newY));
+ }
+ }
+ return res;
+ }
+}
```
### **...**
diff --git a/solution/0700-0799/0778.Swim in Rising Water/README_EN.md b/solution/0700-0799/0778.Swim in Rising Water/README_EN.md
index 881fd72314ff3..1e80146627352 100644
--- a/solution/0700-0799/0778.Swim in Rising Water/README_EN.md
+++ b/solution/0700-0799/0778.Swim in Rising Water/README_EN.md
@@ -102,7 +102,56 @@ We need to wait until time 16 so that (0, 0) and (4, 4) are connected.
### **Java**
```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
+ *
+ * 参考这篇文章的第二种解题方法做的
+ *
+ * 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
+ public int swimInWater(int[][] grid) {
+ // 定义一个优先级队列 按照h从小到大排列
+ Queue>> 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> visitSet = new HashSet<>();
+ visitSet.add(new Pair<>(0, 0));
+
+ int res = 0;
+ int length = grid.length;
+
+ while (!queue.isEmpty()) {
+ Pair> 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;
+ }
+
+ 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;
+ }
+ queue.add(new Pair<>(grid[newX][newY], new Pair<>(newX, newY)));
+ visitSet.add(new Pair<>(newX, newY));
+ }
+ }
+ return res;
+ }
+}
```
### **...**