【LeetCode】200. Number of Islands (2 solutions)

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 @mithmatt for adding this problem and creating all test cases.

 

對每次出現'1'的區域進行計數，同時深度或廣度遍歷，然后置為'0'。

解法一：非遞歸dfs

struct Node
{
    int x;
    int y;
    Node(int newx, int newy): x(newx), y(newy) {}
};

class Solution {
public:
    int numIslands(vector<vector<char>> &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<vector<char>> &grid, int i, int j, int m, int n)
    {
        stack<Node*> 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：遞歸dfs

class Solution {
public:
    int numIslands(vector<vector<char>> &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<vector<char>> &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);
    }
};