827. Making A Large Island-优快云博客

本文链接：https://blog.youkuaiyun.com/zjucor/article/details/80142136

本文介绍了一种算法，用于计算二维网格中通过改变一个0到1所能得到的最大岛屿面积。算法首先标记所有连通的1区域，并为每个连通区域分配一个唯一的ID。然后遍历所有的0，检查周围的1区域ID，计算包含当前0在内的岛屿的最大可能面积。

In a 2D grid of 0s and 1s, we change at most one 0 to a 1.

After, what is the size of the largest island? (An island is a 4-directionally connected group of 1s).

Example 1:

Input: [[1, 0], [0, 1]]
Output: 3
Explanation: Change one 0 to 1 and connect two 1s, then we get an island with area = 3.

Example 2:

Input: [[1, 1], [1, 0]]
Output: 4
Explanation: Change the 0 to 1 and make the island bigger, only one island with area = 1.

Example 3:

Input: [[1, 1], [1, 1]]
Output: 4
Explanation: Can't change any 0 to 1, only one island with area = 1.

Notes:

1 <= grid.length = grid[0].length <= 50.
0 <= grid[i][j] <= 1.

BFS跑一遍，记录每个group的ID

class Solution(object):
    def largestIsland(self, a):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        if not a or not a[0]: return 0
        marked = [[False for _ in range(len(a[0]))] for _ in range(len(a[0]))]
        dirs=[(-1,0),(1,0),(0,1),(0,-1)]
        step = 2
        d={}
        
        for i in range(len(a)):
            for j in range(len(a[0])):
                if not marked[i][j] and a[i][j]==1:
                    marked[i][j]=True
                    q=[(i,j)]
                    cnt=0
                    while q:
                        ii,jj=q.pop()
                        a[ii][jj] = step
                        cnt+=1
                        for di,dj in dirs:
                            iii,jjj=ii+di,jj+dj
                            if 0<=iii<len(a) and 0<=jjj<len(a[0]) and not marked[iii][jjj] and a[iii][jjj]==1:
                                marked[iii][jjj] = True
                                q.append((iii,jjj))
                    d[step]=cnt
                    step+=1
        
        res=max(d.values()) if d else 0
        for i in range(len(a)):
            for j in range(len(a[0])):
                if a[i][j]==0:
                    s=set()
                    for di,dj in dirs:
                        ii,jj=i+di,j+dj
                        if 0<=ii<len(a) and 0<=jj<len(a[0]) and a[ii][jj]!=0:
                            s.add(a[ii][jj])
                    tmp=0
                    for k in s: tmp+=d[k]
                    res=max(res,tmp+1)
        
        return res

s=Solution()
print(s.largestIsland([[1, 0], [0, 1]]))
print(s.largestIsland([[1, 1], [1, 0]]))
print(s.largestIsland([[1, 1], [1, 1]]))
print(s.largestIsland([[0 ,0], [0 ,0]]))