Leetcode 934. Shortest Bridge

Posted 2021-09-03 SnailTyan

文章作者：Tyan
博客：noahsnail.com  |  CSDN  |  简书

1. Description

2. Solution

**解析：**Version 1，先找到矩阵中第一个1作为起点，然后使用广度优先搜索找到所有相邻的1，即第一个岛，并将所有岛的坐标及更改的0计数保存到队列中，初始计数为0，搜索第一个岛的同时，将各个点对应的值设为2，防止重复搜索。从第一个岛的所有点开始，重新使用广度优先搜索，如果搜索的点值为0，将值设为2，表示已经搜索过，同时将点的坐标及计数保存，计数要加1，如果搜索的点为1，说明找到了第二个岛，返回反转的0的计数。

• Version 1

class Solution:
    def shortestBridge(self, grid: List[List[int]]) -> int:
        n = len(grid)
        queue = collections.deque()
        queue2 = collections.deque()
        for i in range(n):
            flag = False
            for j in range(n):
                if grid[i][j] == 1:
                    grid[i][j] = 2
                    queue.append((i, j))
                    flag = True
                    break
            if flag:
                break
        while queue:
            x, y = queue.popleft()
            queue2.append((x, y, 0))
            if x > 0 and grid[x-1][y] == 1:
                grid[x-1][y] = 2
                queue.append((x-1, y))
            if y > 0 and grid[x][y-1] == 1:
                grid[x][y-1] = 2
                queue.append((x, y-1))
            if x < n-1 and grid[x+1][y] == 1:
                grid[x+1][y] = 2
                queue.append((x+1, y))
            if y < n-1 and grid[x][y+1] == 1:
                grid[x][y+1] = 2
                queue.append((x, y+1))
        while queue2:
            x, y, count = queue2.popleft()
            if x > 0:
                if grid[x-1][y] == 0:
                    grid[x-1][y] = 2
                    queue2.append((x-1, y, count + 1))
                elif grid[x-1][y] == 1:
                    return count
            if y > 0:
                if grid[x][y-1] == 0:
                    grid[x][y-1] = 2
                    queue2.append((x, y-1, count + 1))
                elif grid[x][y-1] == 1:
                    return count
            if x < n-1:
                if grid[x+1][y] == 0:
                    grid[x+1][y] = 2
                    queue2.append((x+1, y, count + 1))
                elif grid[x+1][y] == 1:
                    return count
            if y < n-1:
                if grid[x][y+1] == 0:
                    grid[x][y+1] = 2
                    queue2.append((x, y+1, count + 1))
                elif grid[x][y+1] == 1:
                    return count

Reference

1. https://leetcode.com/problems/shortest-bridge/