标签：toggle [] -o body info solution its lte ret
题目描述：
给一个01矩阵，求不同的岛屿的个数。
0代表海，1代表岛，如果两个1相邻，那么这两个1属于同一个岛。我们只考虑上下左右为相邻。
在矩阵：
[
[1, 1, 0, 0, 0],
[0, 1, 0, 0, 1],
[0, 0, 0, 1, 1],
[0, 0, 0, 0, 0],
[0, 0, 0, 0, 1]
]
中有 3 个岛.
题目分析：
循环2维数组，找到其值为1的元素，count++，
然后递归改变其上下左右为1的元素值为0，
循环继续。
源码：
class Solution:
# @param {boolean[][]} grid a boolean 2D matrix
# @return {int} an integer
def numIslands(self, grid):
# Write your code here
if grid is None: return None
if grid == []: return 0
# 当数组不为空时，计算行数和列数
self.n = len(grid)
self.m = len(grid[0])
x = 0
for i in range(self.n):
for j in range(self.m):
if grid[i][j] == 1:
x += 1
grid = self.change(grid,i,j)
return x
def change(self,grid,i,j):
grid[i][j] = 0
if i > 0 and grid[i-1][j] == 1:
# 置当前点上边的点为0
grid = self.change(grid,i-1,j)
if i < self.n-1 and grid[i+1][j] == 1:
# 置当前点下边的点为0
grid = self.change(grid,i+1,j)
if j < self.m-1 and grid[i][j+1] == 1:
# 置当前点右方的点为0
grid = self.change(grid,i,j+1)
if j > 0 and grid[i][j-1] == 1:
# 置当前点左方的点为0
grid = self.change(grid,i,j-1)
return grid
LintCode Python 简单级题目 433.岛屿的个数
标签：toggle [] -o body info solution its lte ret
原文地址：http://www.cnblogs.com/bozhou/p/6956382.html
