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

Credits:

Special thanks to  for adding this problem and creating all test cases.

 

对每次出现'1'的区域进行计数，同时深度或广度遍历，然后置为'0'。

解法一：非递归dfs

struct Node{    int x;    int y;    Node(int newx, int newy): x(newx), y(newy) {}};class Solution {public:    int numIslands(vector
     
      
       > &grid) {        int ret = 0;        if(grid.empty() || grid[0].empty())            return ret;        int m = grid.size();        int n = grid[0].size();        for(int i = 0; i < m; i ++)        {            for(int j = 0; j < n; j ++)            {                if(grid[i][j] == '1')                {                    dfs(grid, i, j, m, n);                    ret ++;                }            }        }        return ret;    }        void dfs(vector
       
        
         > &grid, int i, int j, int m, int n)    {        stack
         
           stk;        Node* rootnode = new Node(i, j);        grid[i][j] = '0';        stk.push(rootnode);        while(!stk.empty())        {            Node* top = stk.top();            if(top->x > 0 && grid[top->x-1][top->y] == '1')            {
    //check up                grid[top->x-1][top->y] = '0';                Node* upnode = new Node(top->x-1, top->y);                stk.push(upnode);                continue;            }            if(top->x < m-1 && grid[top->x+1][top->y] == '1')            {
    //check down                grid[top->x+1][top->y] = '0';                Node* downnode = new Node(top->x+1, top->y);                stk.push(downnode);                continue;            }            if(top->y > 0 && grid[top->x][top->y-1] == '1')            {
    //check left                grid[top->x][top->y-1] = '0';                Node* leftnode = new Node(top->x, top->y-1);                stk.push(leftnode);                continue;            }            if(top->y < n-1 && grid[top->x][top->y+1] == '1')            {
    //check right                grid[top->x][top->y+1] = '0';                Node* rightnode = new Node(top->x, top->y+1);                stk.push(rightnode);                continue;            }            stk.pop();        }    }};
         
        
       
      
     

 

解法二：递归dfs

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