[Leetcode search deep and wide search] islands largest area (695)

topic

Given a two-dimensional array of non-null grid number 0 and 1, is an island formed of a combination of four directions (horizontal or vertical) 1 (representing the land). You can assume that the four edges of the two-dimensional matrix are surrounded by water.

Find a given two-dimensional array in the largest island in the area. (If there are no islands, the area is returned to 0.)

Example 1:

[[0,0,1,0,0,0,0,1,0,0,0,0,0],
 [0,0,0,0,0,0,0,1,1,1,0,0,0],
 [0,1,1,0,1,0,0,0,0,0,0,0,0],
 [0,1,0,0,1,1,0,0,1,0,1,0,0],
 [0,1,0,0,1,1,0,0,1,1,1,0,0],
 [0,0,0,0,0,0,0,0,0,0,1,0,0],
 [0,0,0,0,0,0,0,1,1,1,0,0,0],
 [0,0,0,0,0,0,0,1,1,0,0,0,0]]

For the above should return the given matrix 6. Note that the answer should not be 11, since the islands comprise only four horizontal or vertical direction is '1'.

Example 2:

[[0,0,0,0,0,0,0,0]]

For the above given matrix, 0 is returned.

Note: The length and width of the given grid matrix is ​​not more than 50.

answer

A typical search topic , depth-first and breadth-first can be. (Relatively simple, the idea is not recorded)

1, depth-first DFS, recursively. Time: O (mn), Space: O (mn) 2, breadth-first bfs, with the queue. Time: O (mn), Space: O (mn)

# 深搜

class Solution:
    def __init__(self):
        self.a = []
        self.book = []
        self.next = [
            [0, 1],
            [1, 0],
            [0, -1],
            [-1, 0]
        ]
        self.m = 0
        self.n = 0
        self.max = 0

    def maxAreaOfIsland(self, grid) -> int:
        if not grid:
            return 0
        self.a, self.m, self.n = grid, len(grid), len(grid[0])
        self.book = [
            [0 for _ in range(self.n)] for _ in range(self.m)
        ]
        max = 0
        for i in range(self.m):
            for j in range(self.n):
                if self.a[i][j] == 1:
                    self.max = 0
                    self.book[i][j] = 1
                    self.max += 1
                    self.dfs(i, j)
                    if self.max > max:
                        max = self.max
        return max

    def dfs(self, x, y):
        for i in range(4):
            tx = x + self.next[i][0]
            ty = y + self.next[i][1]
            if tx >= 0 and ty >= 0 and tx < self.m and ty < self.n and self.book[tx][ty] == 0 and self.a[tx][ty] == 1:
                self.max += 1
                self.book[tx][ty] = 1
                self.dfs(tx, ty)
                # 不取消标记，只走一次即可


s = Solution()
ans = s.maxAreaOfIsland([[0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
                         [0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0],
                         [0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0],
                         [0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0],
                         [0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0],
                         [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0],
                         [0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0],
                         [0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0]])
print(ans)
# 6