LeetCode200. Number of Islands

题目：

Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

11110
11010
11000
00000

Answer: 1

Example 2:

11000
11000
00100
00011

Answer: 3

题目分析：

题目意思是将所有相邻的“1”当成一个岛，给定一个二维数组，要我们得出总共有多少个岛

由相邻的“1”构成一个岛我们可以想到只要知道其中一个“1”的所在，就可以使用dfs或者bfs来得到所有相邻的其他的“1”，只要当要扩增的位置都没法扩增（也即没有相邻的“1”）的时候让结果+1即可。至于要知道其中一个“1”的所在我们可以遍历这个二维数组，找出其中没有被浏览过的位置（也即之前的扩增不包括的位置）并扩增即可。

下面是使用dfs的c++代码

C++代码

class Solution {public:    void DFS(vector<vector<char>>& grid, vector<bool>&visit, int i, int j) {        int coloum = grid[0].size();        if (i < 0||j < 0||i >= grid.size()||j >= grid[0].size()) {            return;        }        if (visit[i * coloum + j] == true||grid[i][j] == '0') return;            visit[i * coloum + j] = true;            DFS(grid, visit, i - 1, j);            DFS(grid, visit, i + 1, j);            DFS(grid, visit, i, j - 1);            DFS(grid, visit, i, j + 1);    }    int numIslands(vector<vector<char>>& grid) {        if (grid.size() == 0) return 0;        int row = grid.size(), coloum = grid[0].size();        vector<bool> visit;        for (int i = 0; i < row * coloum; i++) {            visit.push_back(false);        }        int result = 0;        for (int i = 0; i < row; i++) {            for (int j = 0; j < coloum; j++) {                if (visit[i * coloum + j] == false&&grid[i][j] == '1') {                    DFS(grid, visit, i, j);                    result++;                }            }        }        return result;    }};