C++递归算法处理岛屿问题详解

    public void dfs(char[][] grid, int m, int n){
    	// 位置越界 或者 该位置已经被遍历过
        if(isBeyond(grid, m, n) || grid[m][n] == 2){
            return;
        }
        // 相应操作
        ........
        // 记录已经遍历过位置
        grid[m][n] = '2';
		// 递归遍历该陆地位置的上下左右位置
        dfs(grid, m-1, n);
        dfs(grid, m+1, n);
        dfs(grid, m, n-1);
        dfs(grid, m, n+1);
    }
	// 检测越界的函数
    boolean isBeyond(char[][] grid, int m, int n){
        if(m < 0 || m>=grid.length || n<0 || n>=grid[0].length){
            return true;
        }
        return false;
    }

class Solution {
    public int numIslands(char[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        int ans = 0;
        for(int i=0; i<m; i++){
            for(int j=0; j<n;j++){
                if(grid[i][j] == '1'){
                    // 递归过程中将属于同一岛屿的位置标记为2，保证属于同一岛屿的陆地1位置不会重复进入循环
                    dfs(grid, i, j); 
                    ans++;
                }
            }
        }
        return ans;
    }
    public void dfs(char[][] grid, int m, int n){
        if(isBeyond(grid, m, n)){
            return;
        }
        if(grid[m][n] != '1'){
            return;
        }
        grid[m][n] = '2';
        dfs(grid, m-1, n);
        dfs(grid, m+1, n);
        dfs(grid, m, n-1);
        dfs(grid, m, n+1);
    }
    boolean isBeyond(char[][] grid, int m, int n){
        if(m < 0 || m>=grid.length || n<0 || n>=grid[0].length){
            return true;
        }
        return false;
    }
}

class Solution {
    // 从一个陆地方块走向一个非陆地方块，就将岛屿面积加1
    public int islandPerimeter(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        int perimeter = 0;
        for(int i=0; i<m; i++){
            for(int j=0; j<n; j++){
                if(grid[i][j] == 1){
                    // 因为只有一个岛屿，直接返回即可
                    return getPerimeter(grid, i, j);
                }
            }
        }
        return 0;
    }
    public int getPerimeter(int[][] grid, int m, int n){
        // 走到非陆地方块，返回共享度1
        if(isBeyond(grid, m, n) || grid[m][n] == 0){
            return 1;
        }
        // 走到遍历过方块返回0
        if(grid[m][n] == 2){
            return 0;
        }
        // 标记已经遍历过节点
        grid[m][n] = 2;
        return getPerimeter(grid, m-1, n)
        + getPerimeter(grid, m+1, n)
        + getPerimeter(grid, m, n-1)
        + getPerimeter(grid, m, n+1);
    }
    boolean isBeyond(int[][] grid, int m, int n){
        if(m < 0 || n < 0 || m >= grid.length || n >= grid[0].length){
            return true;
        }
        return false;
    }
}