力扣题/图论/岛屿数量

岛屿数量

力扣原题

给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。

岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。

此外，你可以假设该网格的四条边均被水包围。

• 示例 1:

输入：grid = [

"1","1","1","1","0"\], \["1","1","0","1","0"\], \["1","1","0","0","0"\], \["0","0","0","0","0"

]

输出：1

• 示例 2:

输入：grid = [

"1","1","0","0","0"\], \["1","1","0","0","0"\], \["0","0","1","0","0"\], \["0","0","0","1","1"

]

输出：3

/**
     * @param {character[][]} grid
     * @return {number}
     */
var numIslands = function (grid) {
    const m = grid.length
    const n = grid[0].length
    let landCount = 0;

    function expandLand(row, col) {
        const isLand = grid[row][col]
        if (isLand !== '1') return
        grid[row][col] = '-1' // 访问过的土地，不可再访问
        if (row > 0) {
            // 上
            expandLand(row - 1, col)
        }
        if (row < m - 1) {
            // 下
            expandLand(row + 1, col)
        }
        if (col > 0) {
            // 左
            expandLand(row, col - 1)
        }
        if (col < n - 1) {
            // 右
            expandLand(row, col + 1)
        }
    }

    for (let r = 0; r < m; r++) {
        for (let c = 0; c < n; c++) {
            // 如果当前位置是陆地，则为新陆地
            const isLand = grid[r][c]
            // 访问周边陆地
            if (isLand === '1') {
                landCount += 1
                expandLand(r, c) // 扩张联通的陆地，全部置为已访问
            }
        }
    }

    return landCount
};

解题思路

遍历所有位置，首次遇到值为'1'陆地 时，此时岛屿数量+1，然后从上下左右四个方向寻找相连通的陆地，并标识为'-1'已访问状态，再寻找下一个陆地就不会重复了