200. 岛屿数量

题号：200
来源：LeetCode
难度：中等
标签：DFS 图 矩阵
语言：TypeScript
解法：1 个
作者：lmliheng
最近更新：2026-09-27

TypeScript · O(m*n) 时间 · O(m*n) 空间 · 更新于 2026-09-27

DFS 淹没法，统计连通块数量

typescript

/**
 * @difficulty medium
 * @tags DFS,图,矩阵
 * @time O(m*n)
 * @space O(m*n)
 * @note DFS 淹没法，统计连通块数量
 * @200. 岛屿数量
 */

let grid: string[][] = [
["1","0","1","1","1"],
["1","0","1","0","1"],
["1","1","1","0","1"]]

let res: number = 0
let m: number = grid.length
let n: number = grid[0].length
console.log(m,n)
const dfs = (i: number, j: number): void => {
    if (i < 0 || i >= m || j < 0 || j >= n) {
        return
    }

    if(grid[i][j]!=='1'){
        return
    }
    console.log('成功遍历的节点：',i,j)
    grid[i][j] = '2'
    dfs(i + 1, j)
    dfs(i, j + 1)
    dfs(i , j-1)
    dfs(i-1 , j)
    // dfs(i + 1, j + 1)

}

for (let i: number = 0; i < m; i++) {
    for (let j: number = 0; j < n; j++) {
        if (grid[i][j] === '1') {
            dfs(i, j)
            console.log('dfs起点',i,j)
            res++

        }
    }
}

console.log(res)
console.log(grid)

export {};

源码：ts/leetcode/200. 岛屿数量.ts

在 GitHub 上查看题目所在目录：lmliheng/algorithm

200. 岛屿数量 ​

200. 岛屿数量