題目如下：

解題思路：這個題目可以進行拆分成幾個子問題。第一，求出island的數量，其實就是 200. Number of Islands，這個很簡單，DFS或者BFS都能搞定；第二，除了求出island的數量之外，還要求出每個island包括的1的數量，這個也不難，在DFS或者BFS的過程中計數即可；第三，遍歷grid中所有的0，判斷每個0的上下左右分別連接了幾個不同的island，並將連接的所有island的所有1的數量求和，再加上本身的1（0變來的）即是這個0變成1可以得到的large island，最後，求出所有0的最大的large island即可。

代碼如下：

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為island的編號，用來記錄一個有幾個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為每個island的1的數量
                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

【leetcode】827. Making A Large Island