原题链接：1254. 统计封闭岛屿的数目 - 力扣（LeetCode） (leetcode-cn.com)
考察知识点： dfs
解题思路与源代码：
/**
* @param {number[][]} grid
* @return {number}
*/
var closedIsland = function(grid) {
// isBorder为是否超出边界 , resCount为封闭岛屿数量
let isBorder = 0, resCount = 0;
// 判断是否超界
const isArea = (grid, row, col) => {
if(row<0 || row>=grid.length || col<0 || col>=grid[0].length){
return false
}
return true;
}
//dfs深度优先算法
const dfsArea = (grid, row, col)=> {
if (!isArea(grid, row, col)){
isBorder++;
return;
}
//如果不是陆地或者已经遍历过停止该方向遍历
if(grid[row][col] != 0){
return;
}
//遍历过后，记录已遍历标记为2
grid[row][col] = 2;
//四个方向进行遍历
dfsArea(grid, row-1, col);
dfsArea(grid, row, col+1);
dfsArea(grid, row+1, col);
dfsArea(grid, row, col-1);
}
for (let i=0; i<grid.length; i++){
for (let j=0; j<grid[0].length; j++){
if (grid[i][j] === 0){
dfsArea(grid, i, j);
if (isBorder === 0){
resCount++;
}
isBorder = 0;
}
}
}
return resCount;
};
