leetcode827. Making A Large Island

Posted 2020-11-01 seyjs

题目如下：

解题思路：这个题目可以进行拆分成几个子问题。第一，求出island的数量，其实就是 200. Number of Islands，这个很简单，DFS或者BFS都能搞定；第二，除了求出island的数量之外，还要求出每个island包括的1的数量，这个也不难，在DFS或者BFS的过程中计数即可；第三，遍历grid中所有的0，判断每个0的上下左右分别连接了几个不同的island，并将连接的所有island的所有1的数量求和，再加上本身的1（0变来的）即是这个0变成1可以得到的large island，最后，求出所有0的最大的large island即可。

代码如下：

class Solution(object):
    def largestIsland(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        for i in grid:
            i.append(\'#\')
            i.insert(0,\'#\')
        grid.append([\'#\' for i in grid[0]])
        grid.insert(0,[\'#\' for i in grid[0]])
        visit = []
        for i in grid:
            visit.append([0 for x in i])
        queue = []

        l = [0]

        area = 1
        num = 1

        for i in xrange(1,len(grid)):
            for j in xrange(1,len(grid[i])):
                if grid[i][j] == \'#\' or grid[i][j] == 0:
                    continue
                if visit[i][j] != 0:
                    continue
                queue.append((i,j,area))
                visit[i][j] = area
                while len(queue) > 0:
                    x,y,a = queue.pop(0) # a为island的编号，用来记录一个有几个island
                    if grid[x+1][y] == 1 and visit[x+1][y] == 0:
                        num += 1
                        queue.append((x+1,y,a))
                        visit[x + 1][y] = a
                    if grid[x-1][y] == 1 and visit[x-1][y] == 0:
                        num += 1
                        queue.append((x-1,y,a))
                        visit[x - 1][y] = a
                    if grid[x][y+1] == 1 and visit[x][y+1] == 0:
                        num += 1
                        queue.append((x,y+1,a))
                        visit[x][y + 1] = a
                    if grid[x][y-1] == 1 and visit[x][y-1] == 0:
                        num += 1
                        queue.append((x,y-1,a))
                        visit[x][y - 1] = a
                area += 1
                l.append(num) #l为每个island的1的数量
                num = 1
        res = 0
        for i in l:
            if res < i:
                res = i
        #print visit,l
        for i in xrange(1,len(grid)):
            for j in xrange(1,len(grid[i])):
                if grid[i][j] == 0:
                    count = 1
                    al = []
                    if grid[i+1][j] == 1:
                        if visit[i+1][j] not in al:
                            count += l[visit[i+1][j]]
                            al.append(visit[i+1][j])
                    if grid[i-1][j] == 1:
                        if visit[i-1][j] not in al:
                            count += l[visit[i-1][j]]
                            al.append(visit[i-1][j])
                    if grid[i][j+1] == 1:
                        if visit[i][j+1] not in al:
                            count += l[visit[i][j+1]]
                            al.append(visit[i][j+1])
                    if grid[i][j-1] == 1:
                        if visit[i][j-1] not in al:
                            count += l[visit[i][j-1]]
                            al.append(visit[i][j-1])
                    if res < count:
                        res = count
        return res