200. Number of Islands 岛屿数量 (C++和Java解法)

Input: grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]
Output: 3

题目大意：1代表土地，0代表水，一个岛屿是由四面环水的土地组成，求岛屿的数量。如上例中含3个岛屿。

解题思路:

典型的回溯法(DFS)求解。

DFS逻辑：目的是搜索完某个岛屿所有为1的所有区域。访问标志visited[i][j]设为true，再对grid[i][j]的上下左右区域进行递归

DFS结束条件：遇到0或者超出边界

C++

class Solution {
public:
    int result = 0;
    int m, n;
    
    vector<vector<bool>> visited;
    
    void DFS(int i, int j, vector<vector<char>>& grid){
        if(i >= 0 && j >= 0 && i < m && j < n && !visited[i][j] && grid[i][j] == '1'){
            visited[i][j] = true;
            DFS(i - 1, j, grid);
            DFS(i + 1, j, grid);
            DFS(i, j + 1, grid);
            DFS(i, j - 1, grid);
        }
    }
    
    void init(){
        for(int i = 0;i < m;i++){
            visited.push_back(vector<bool>(n, false));
        }
    }
    
    int numIslands(vector<vector<char>>& grid) {
        if(grid.size() == 0 || grid[0].size() == 0){
            return 0;
        }
        m = grid.size();
        n = grid[0].size();
        init();
        for(int i = 0;i < m;i++){
            for(int j = 0;j < n;j++){
                if(!visited[i][j] && grid[i][j] == '1'){
                    //找到一个岛屿，并搜索完该岛屿所有区域，岛屿数量增加
                    DFS(i, j, grid);
                    result++;
                }        
            }
        }
        return result;
    }
};

Java

class Solution {
    int m, n;
    int result = 0;
    boolean[][] visited;
    
    void DFS(int i, int j, char[][] grid){
        if(i >= 0 && j >= 0 && i < m && j < n && !visited[i][j] && grid[i][j] == '1'){
            visited[i][j] = true;
            DFS(i - 1, j, grid);
            DFS(i + 1, j, grid);
            DFS(i, j + 1, grid);
            DFS(i, j - 1, grid);
        }
    }
    
    public int numIslands(char[][] grid) {
        if(grid.length == 0 || grid[0].length == 0){
            return 0;
        }
        m = grid.length;
        n = grid[0].length;
        visited = new boolean[m][n];
        
        for(int i = 0;i < m;i++){
            for(int j = 0;j < n;j++){
                if(!visited[i][j] && grid[i][j] == '1'){
                    DFS(i, j, grid);
                    result++;
                }        
            }
        }
        return result;
    }
}

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