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题目

基本思路

遍历矩阵，每遇到’1’后, 开始向四个方向递归搜索，搜到后变为’0’，因为相邻的属于一个island，然后开始继续找下一个’1’。

实现代码

class Solution:    def numIslands(self, grid):        """        :type grid: List[List[str]]        :rtype: int        """        m = len(grid)        if m == 0:            return 0        n = len(grid[0])        if n == 0:            return 0        # 结果        count = 0        for i in range(m):            for j in range(n):                if grid[i][j] == "1":                    self.islandErase(grid,i,j)                    count += 1        return count        def islandErase(self,grid,i,j):        if i<0 or j<0 or i>=len(grid) or j>=len(grid[0]) or grid[i][j]=="0":            return                 grid[i][j] = "0"        self.islandErase(grid,i-1,j)        self.islandErase(grid,i+1,j)        self.islandErase(grid,i,j-1)        self.islandErase(grid,i,j+1)