[leetcode] 200. Number of Islands

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

这道题是统计小岛的个数，题目难度为Medium。

针对grid[i][j]，如果是‘1’，表明已经进入一个小岛，将小岛数加一并把grid[i][j]修改为‘0’以表示这个点已统计过，然后依次查看和它相邻的四个点，如果是‘1’，将其变为‘0’，因为这个点和grid[i][j]在一个岛上，然后再依次搜索和他们相邻的点，直至所有相邻的点都是‘0’，这样一个完整的小岛就遍历完了，所有的点都从‘1’变为了‘0’。以上方法即所谓的深度优先搜索策略。具体代码：

class Solution {    void reduceSameIsland(vector<vector<char>>& grid, int i, int j) {        if(i<0 || i>=grid.size() || j<0 || j>=grid[i].size() || grid[i][j]=='0') return;        grid[i][j] = '0';        reduceSameIsland(grid, i-1, j);        reduceSameIsland(grid, i+1, j);        reduceSameIsland(grid, i, j-1);        reduceSameIsland(grid, i, j+1);    }public:    int numIslands(vector<vector<char>>& grid) {        int cnt = 0;        for(int i=0; i<grid.size(); i++) {            for(int j=0; j<grid[i].size(); j++) {                if(grid[i][j] == '1') {                    cnt++;                    reduceSameIsland(grid, i, j);                }            }        }        return cnt;    }};

还可以通过广度优先的策略进行处理，针对grid[i][j]，如果为‘1’，将其修改为‘0’，然后将和它相邻的四个点中为‘1’的点存入队列，这样依次遍历，直至队列为空，一个小岛就处理完了。具体代码：

class Solution {    void reduceSameIsland(vector<vector<char>>& grid, int i, int j) {        queue<int> idx;        int col = grid[0].size();        grid[i][j] = '0';        idx.push(i*col+j);        while(!idx.empty()) {            int sz = idx.size();            for(int k=0; k<sz; k++) {                int index = idx.front();                int m = index/col;                int n = index%col;                idx.pop();                if(m-1>=0 && grid[m-1][n]=='1') {                    grid[m-1][n] = '0';                    idx.push(index-col);                }                if(m+1<grid.size() && grid[m+1][n]=='1') {                    grid[m+1][n] = '0';                    idx.push(index+col);                }                if(n-1>=0 && grid[m][n-1]=='1') {                    grid[m][n-1] = '0';                    idx.push(index-1);                }                if(n+1<col && grid[m][n+1]=='1') {                    grid[m][n+1] = '0';                    idx.push(index+1);                }            }        }    }public:    int numIslands(vector<vector<char>>& grid) {        int cnt = 0;        for(int i=0; i<grid.size(); i++) {            for(int j=0; j<grid[0].size(); j++) {                if(grid[i][j] == '1') {                    cnt++;                    reduceSameIsland(grid, i, j);                }            }        }        return cnt;    }};