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// @Title: 岛屿数量 (Number of Islands)
// @Author: 15816537946@163.com
// @Date: 2020-06-25 07:15:36
// @Runtime: 0 ms
// @Memory: 2.9 MB
/*
 * @lc app=leetcode.cn id=200 lang=golang
 *
 * [200] 岛屿数量
 *
 * https://leetcode-cn.com/problems/number-of-islands/description/
 *
 * algorithms
 * Medium (45.46%)
 * Likes:    222
 * Dislikes: 0
 * Total Accepted:    26.5K
 * Total Submissions: 58.4K
 * Testcase Example:  '[["1","1","1","1","0"],["1","1","0","1","0"],["1","1","0","0","0"],["0","0","0","0","0"]]'
 *
 * 给定一个由 '1'(陆地)和
 * '0'(水)组成的的二维网格,计算岛屿的数量。一个岛被水包围,并且它是通过水平方向或垂直方向上相邻的陆地连接而成的。你可以假设网格的四个边均被水包围。
 *
 * 示例 1:
 *
 * 输入:
 * 11110
 * 11010
 * 11000
 * 00000
 *
 * 输出: 1
 *
 *
 * 示例 2:
 *
 * 输入:
 * 11000
 * 11000
 * 00100
 * 00011
 *
 * 输出: 3
 *
 *
 */
 /*
func numIslands(grid [][]byte) int {
	n := len(grid)
	if n == 0 {
		return 0
	}
	m := len(grid[0])

	var dfs func(int, int)
	dfs = func(i, j int) {
		grid[i][j] = '0'

		if 0 <= i-1 && grid[i-1][j] == '1' {
			dfs(i-1, j)
		}
		if 0 <= j-1 && grid[i][j-1] == '1' {
			dfs(i, j-1)
		}
		if i+1 < n && grid[i+1][j] == '1' {
			dfs(i+1, j)
		}
		if j+1 < m && grid[i][j+1] == '1' {
			dfs(i, j+1)
		}
	}

	var ret int
	for i := range grid {
		for j := range grid[0] {
			if grid[i][j] == '1' {
				ret++
				dfs(i, j)
			}
		}
	}
	return ret
}
*/
func numIslands(grid [][]byte) int {
	n := len(grid)
	if n == 0 {
		return 0
	}
	m := len(grid[0])

	var dfs func(int, int)
	dfs = func(i, j int) {
		grid[i][j] = '0'

		if i > 0 && grid[i-1][j] == '1' {
			dfs(i-1, j)
		}
		if j > 0 && grid[i][j-1] == '1' {
			dfs(i, j-1)
		}
		if i < n-1 && grid[i+1][j] == '1' {
			dfs(i+1, j)
		}
		if j < m-1 && grid[i][j+1] == '1' {
			dfs(i, j+1)
		}
	}

	var cnt int
	for i :=range grid {
		for j := range grid[0] {
			if grid[i][j] == '1' {
				cnt++
				dfs(i, j)
			}
		}
	}

	return cnt

}