【广度优先遍历-中等】695. 岛屿的最大面积、剑指 Offer II 105. 岛屿的最大面积

【题目】

【代码】
其实这题的本质就是：岛屿数量
而岛屿数量的本质就是：非联通子图的个数
岛屿这类题目都是以矩阵的形式给出的，矩阵的本质和无向图是一样的，都需要记录是否访问过某个点
这题求最大的岛屿面积，首先需要遍历每个点，判断该点是否被访问过，如果该点是岛屿的一部分且没有被访问过，则遍历该岛屿的面积

class Solution:
    def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
        visit=[[0 for i in range(len(grid[0]))] for i in range(len(grid))]
        ans=0
        def dfs(x,y):
            if x>=0 and x<len(grid) and y>=0 and y<len(grid[0]) and grid[x][y]==1 and visit[x][y]==0:
                visit[x][y]=1
                return 1+dfs(x-1,y)+dfs(x+1,y)+dfs(x,y+1)+dfs(x,y-1)
            return 0
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j]==1 and visit[i][j]==0:
                    ans=max(ans,dfs(i,j))
        return ans

【方法2】不占用额外空间 不用单独设置visited数组

class Solution:
    def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
        ans=0
        def dfs(x,y):
            if x>=0 and x<len(grid) and y>=0 and y<len(grid[0]) and grid[x][y]==1:
                grid[x][y]=0
                return 1+dfs(x-1,y)+dfs(x+1,y)+dfs(x,y+1)+dfs(x,y-1)
            return 0
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j]==1:
                    ans=max(ans,dfs(i,j))
        return ans

【方法3】栈代替递归

class Solution:
    def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
        ans=0
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j]==1:
                    cnt=0
                    stack=[(i,j)]
                    while stack:
                        x,y=stack.pop()
                        if x>=0 and x<len(grid) and y>=0 and y<len(grid[0]) and grid[x][y]==1:
                            cnt+=1
                            grid[x][y]=0
                            stack.append((x-1,y))
                            stack.append((x+1,y))
                            stack.append((x,y+1))
                            stack.append((x,y-1))
                    ans=max(ans,cnt)
        return ans