【leetcode】剑指 Offer II 105. 岛屿的最大面积-【深度优先DFS】

m == grid.length
    n == grid[i].length
    1 <= m, n <= 50
    grid[i][j] is either 0 or 1

/**
 * @param {number[][]} grid
 * @return {number}
 */
var maxAreaOfIsland = function (grid) {
    let col = grid.length,row = grid[0].length,max=0; //行、列
    //未访问过的岛屿为1，访问过的岛屿为0
    let matrix = [...Array(col)].fill(0).map(()=> [...Array(row)].fill(1))
    let dfs = (i,j)=>{
        //边界
        if(i<0 || i>=col || j<0 || j>=row) return 0;
        let ret=0;
        let temp = matrix[i][j] //是否为岛屿
        matrix[i][j] = 0 //访问过的岛屿标记为 0
        if(grid[i][j] && temp){
            ret+=dfs(i+1,j) //下
            ret+=dfs(i-1,j) //上
            ret+=dfs(i,j+1) //右
            ret+=dfs(i,j-1) //左
            ret++
        }
        return ret
    }
    for(let i=0;i<col;i++){
        for(let j=0;j<row;j++){
            if(grid[i][j]===1){
                //更新岛屿记录
                max = Math.max(max,dfs(i,j))
            }
        }
    }
    return max
};