岛屿的个数-lintcode

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此题用到了深度搜索的思想，在depth_search()中，首先判断边界条件和是否为0的点，然后对该点的上下左右进行depth_search()：

class Solution:    # @param {boolean[][]} grid a boolean 2D matrix    # @return {int} an integer    def numIslands(self, grid):        if len(grid) == 0 or len(grid[0]) == 0:            return 0        self.rows = len(grid)        self.clos = len(grid[0])        self.count = 0        for i in range(self.rows):            for j in range(self.clos):                if grid[i][j] == 1:                    self.depth_search(grid, i, j)                    self.count += 1        return self.count    def depth_search(self, grid, i, j):        if i < 0 or i >= self.rows or j < 0 or j >= self.clos:            return        if grid[i][j] == 0:            return        grid[i][j] = 0        if i < self.rows-1:            self.depth_search(grid, i+1, j)        if j < self.clos - 1:            self.depth_search(grid, i, j+1)        if i > 0:            self.depth_search(grid, i-1, j)        if j > 0:            self.depth_search(grid, i, j-1)

C++代码：

class Solution {public:    /**     * @param grid a boolean 2D matrix     * @return an integer     */    int numIslands(vector<vector<bool>>& grid) {        if (grid.size() == 0 || grid[0].size() == 0) {            return 0;        }        int count = 0;        int rows = grid.size();        int clos = grid[0].size();        for (int i =0; i<rows; i++){            for (int j =0; j<clos; j++) {                if (grid[i][j] == true) {                    depth_search(grid, i, j);                    count++;                }            }        }        return count;    }    void depth_search(vector<vector<bool>>& grid, int i, int j) {        if(i<0 || i>grid.size()-1 || j>grid[0].size()-1 || j<0 )            return;        if (grid[i][j] == false)            return;        grid[i][j] = false;        if (i>0)            depth_search(grid, i-1, j);        if (i<grid.size()-1)            depth_search(grid, i+1, j);        if (j>0)            depth_search(grid, i, j-1);        if (j<grid[0].size()-1)            depth_search(grid, i, j+1);    }};

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