leetcode 200. Number of islands

difficulty ： secondary
The frequency of ：122

subject ： Here you are ‘1’（ land ） and ‘0’（ water ） A two-dimensional mesh of , Please calculate the number of islands in the grid .

Islands are always surrounded by water , And each island can only be combined by horizontal direction / Or a vertical connection between adjacent lands .

Besides , You can assume that all four sides of the mesh are surrounded by water .

Their thinking ： Depth-first search

Be careful ：

• DFS It is necessary to judge whether it is still in the matrix ( Jump out of bounds DFS) And whether that point is land （ If the ocean jumps out DFS）
• Depth-first search , You need to avoid repeated iterations （ Mark the passing become 0）
• Two are required in the main function for, Do a two-dimensional traversal
• row=grid.length;col=grid[0].length

Code

class Solution {
    
    public int numIslands(char[][] grid) {
    
       //row  The length of the line  col  Column length 
       int r=grid.length;
       int c=grid[0].length;
       int count=0;
       for(int i=0;i<r;i++){
    
           for(int j=0;j<c;j++){
    
               // If it's land , it DFS, Jump out by the way DFS In time COUNT++
               if(grid[i][j]=='1'){
    
                   DFS(grid,i,j);
                   count++;
               }
           }
       }
       return count;
    }
    // You don't need to return anything , So it's void
    public void DFS(char[][] grid,int r,int c){
    
        // Judge (r,c) Whether in grid in , Out of or for the sea 
        if(r<0||c<0||r>=grid.length||c>=grid[0].length) return;
        // Here are two steps to take apart , If put together , There may be a mistake . Is due to || It's the nature of 
        if(grid[r][c]=='0') return;
        grid[r][c]='0';
        // On 
        DFS(grid,r-1,c);
        // Next 
        DFS(grid,r+1,c);
        // Left 
        DFS(grid,r,c-1);
        // Right 
        DFS(grid,r,c+1);
    }
}