LeetCode: Number of Islands

## https://code.dennyzhang.com/number-of-islands
class Solution(object):
    ## Basic Ideas: Avoid duplicate counting.
    ##   Mark all adjacent 1 to X. Thus we can avoid counting one the same island multiple times.
    ##
    ## Complexity: Time O(m*n), Space O(1)
    def numIslands(self, grid):
        """
        :type grid: List[List[str]]
        :rtype: int
        """
        self.row_count = len(grid)
        if self.row_count == 0: return 0
        self.col_count = len(grid[0])

        res = 0
        for i in xrange(self.row_count):
            for j in xrange(self.col_count):
                if grid[i][j] == '1':
                    res += 1
                    self.DFSMark(grid, i, j)
        return res

    def DFSMark(self, grid, i, j):
        if i < 0 or i >= self.row_count \
            or j < 0 or j >= self.col_count:
            return

        # stop digging, if not '1'
        if grid[i][j] != '1': return

        grid[i][j] = 'X'
        # mark four positions in a recursive way
        self.DFSMark(grid, i-1, j)
        self.DFSMark(grid, i+1, j)
        self.DFSMark(grid, i, j-1)
        self.DFSMark(grid, i, j+1)