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[pull] master from youngyangyang04:master #87

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Merged
pull merged 4 commits into AlgorithmAndLeetCode:master from youngyangyang04:master
Sep 13, 2022
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47 changes: 47 additions & 0 deletions problems/0200.岛屿数量.广搜版.md
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -144,3 +144,50 @@ public:
};

```

## 其他语言版本

### Java

```java
class Solution {

boolean[][] visited;
int[][] move = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

public int numIslands(char[][] grid) {
int res = 0;
visited = new boolean[grid.length][grid[0].length];
for(int i = 0; i < grid.length; i++) {
for(int j = 0; j < grid[0].length; j++) {
if(!visited[i][j] && grid[i][j] == '1') {
bfs(grid, i, j);
res++;
}
}
}
return res;
}

//将这片岛屿上的所有陆地都访问到
public void bfs(char[][] grid, int y, int x) {
Deque<int[]> queue = new ArrayDeque<>();
queue.offer(new int[]{y, x});
visited[y][x] = true;
while(!queue.isEmpty()) {
int[] cur = queue.poll();
int m = cur[0];
int n = cur[1];
for(int i = 0; i < 4; i++) {
int nexty = m + move[i][0];
int nextx = n + move[i][1];
if(nextx < 0 || nexty == grid.length || nexty < 0 || nextx == grid[0].length) continue;
if(!visited[nexty][nextx] && grid[nexty][nextx] == '1') {
queue.offer(new int[]{nexty, nextx});
visited[nexty][nextx] = true; //只要加入队列就标记为访问
}
}
}
}
}
```
77 changes: 76 additions & 1 deletion problems/0695.岛屿的最大面积.md
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -170,6 +170,81 @@ public:

# 其它语言版本

## Python
### BFS
```python
class Solution:
def __init__(self):
self.count = 0

def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
# 与200.独立岛屿不同的是:此题grid列表内是int!!!

# BFS
if not grid: return 0

m, n = len(grid), len(grid[0])
visited = [[False for i in range(n)] for j in range(m)]

result = 0
for i in range(m):
for j in range(n):
if not visited[i][j] and grid[i][j] == 1:
# 每一个新岛屿
self.count = 0
print(f'{self.count}')
self.bfs(grid, visited, i, j)
result = max(result, self.count)

return result

def bfs(self, grid, visited, i, j):
self.count += 1
visited[i][j] = True

queue = collections.deque([(i, j)])
while queue:
x, y = queue.popleft()
for new_x, new_y in [(x + 1, y), (x - 1, y), (x, y - 1), (x, y + 1)]:
if 0 <= new_x < len(grid) and 0 <= new_y < len(grid[0]) and not visited[new_x][new_y] and grid[new_x][new_y] == 1:
visited[new_x][new_y] = True
self.count += 1
queue.append((new_x, new_y))
```
### DFS
```python
class Solution:
def __init__(self):
self.count = 0

def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
# DFS
if not grid: return 0

m, n = len(grid), len(grid[0])
visited = [[False for _ in range(n)] for _ in range(m)]

result = 0
for i in range(m):
for j in range(n):
if not visited[i][j] and grid[i][j] == 1:
self.count = 0
self.dfs(grid, visited, i, j)
result = max(result, self.count)
return result

def dfs(self, grid, visited, x, y):
if visited[x][y] or grid[x][y] == 0:
return
visited[x][y] = True
self.count += 1
for new_x, new_y in [(x + 1, y), (x - 1, y), (x, y + 1), (x, y - 1)]:
if 0 <= new_x < len(grid) and 0 <= new_y < len(grid[0]):
self.dfs(grid, visited, new_x, new_y)
```



## Java

这里使用深度优先搜索 DFS 来完成本道题目。我们使用 DFS 计算一个岛屿的面积,同时维护计算过的最大的岛屿面积。同时,为了避免对岛屿重复计算,我们在 DFS 的时候对岛屿进行 "淹没" 操作,即将岛屿所占的地方置为 0。
Expand Down Expand Up @@ -199,4 +274,4 @@ public int dfs(int[][] grid,int i,int j){
dfs(grid,i,j + 1) +
dfs(grid,i,j - 1);
}
```
```

AltStyle によって変換されたページ (->オリジナル) /