827

Author: ProHiryu

DFS/traditionalDFS/827.md

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+## Making A Large Island
+
+#### Description
+
+[link](https://leetcode.com/problems/making-a-large-island/)
+
+---
+
+#### Solution
+
+- See Code
+
+---
+
+#### Code
+
+> 最坏情况：O(n^2)
+
+```python
+class Solution:
+    '''
+    思路：
+        1. 使用index来表示不同的island
+        2. 更改grid当中表示陆地的值为对应的index
+        3. 先将所有的island面积计算出来，同时保存在对应index的面积中
+        4. 遍历0的值，计算能够连接起来的最大值即可
+    '''
+    def largestIsland(self, grid: List[List[int]]) -> int:
+        N = len(grid)
+        def move(x, y):
+            for i, j in ((1, 0), (-1, 0), (0, 1), (0, -1)):
+                if 0 <= x + i < N and 0 <= y + j < N:
+                    yield x + i, y + j
+
+        def dfs(x, y, index):
+            area = 0
+            grid[x][y] = index
+            for i, j in move(x, y):
+                if grid[i][j] == 1:
+                    area += dfs(i, j, index)
+            return area + 1
+
+        # DFS every island and give it an index of island
+        index = 2
+        area = {}
+        for x in range(N):
+            for y in range(N):
+                if grid[x][y] == 1:
+                    area[index] = dfs(x, y, index)
+                    index += 1
+
+        # traverse every 0 cell and count biggest island it can conntect
+        res = max(area.values() or [0])
+        for x in range(N):
+            for y in range(N):
+                if grid[x][y] == 0:
+                    possible = set(grid[i][j] for i, j in move(x, y) if grid[i][j] > 1)
+                    res = max(res, sum(area[index] for index in possible) + 1)
+        return res
+```