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Merged
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Oct 15, 2020
Merged

add java solution for leetcode 778 #300

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51 changes: 50 additions & 1 deletion solution/0700-0799/0778.Swim in Rising Water/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -64,7 +64,56 @@
<!-- 这里可写当前语言的特殊实现逻辑 -->

```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
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;
}

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;
}
}
```

### **...**
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51 changes: 50 additions & 1 deletion solution/0700-0799/0778.Swim in Rising Water/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -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
* <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
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;
}

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;
}
}
```

### **...**
Expand Down