200. Number of Islands

Given an m x n 2D binary grid grid which represents a map of '1’s (land) and '0’s (water), return the number of islands.

An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

Input: grid = [["1","1","1","1","0"],["1","1","0","1","0"],["1","1","0","0","0"],["0","0","0","0","0"]
]
Output: 1

Example 2:

Input: grid = [["1","1","0","0","0"],["1","1","0","0","0"],["0","0","1","0","0"],["0","0","0","1","1"]
]
Output: 3

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 300
grid[i][j] is '0' or '1'.

1. 递归解法

2层循环遍历是否为’1’, 如果是，则递归循环打掉所有上下左右的的’1’. 再接着遍历。如果还能优化，可以记录遍历过的记录，直接跳过打掉过的1.

class Solution {public int numIslands(char[][] grid) {int count = 0;for (int i = 0; i < grid.length; i++) {for (int k = 0; k < grid[0].length; k++) {if (grid[i][k] == '1') {count++;sink(grid, i, k);}}}return count;}private void sink(char[][] grid, int i, int k) {if (i < 0 || k < 0 || i >= grid.length || k >= grid[0].length || grid[i][k] == '0') return;grid[i][k] = '0';sink(grid, i+1, k);sink(grid, i, k+1);sink(grid, i-1, k);sink(grid, i, k-1);}
}

2. 递归求解，优化掉重复的sink函数

看到优雅的写法就是上下左右，用一个数组{0, 1, 0, -1, 0} 两两遍历，

下 0,1
右1,0
上0, -1
左-1, 0

class Solution {public int numIslands(char[][] grid) {int count = 0;for (int i = 0; i < grid.length; i++) {for (int k = 0; k < grid[0].length; k++) {if (grid[i][k] == '1') {count++;sink(grid, i, k);}}}return count;}int[] d = {0, 1, 0, -1, 0};private void sink(char[][] grid, int i, int k) {if (i < 0 || k < 0 || i >= grid.length || k >= grid[0].length || grid[i][k] == '0') return;grid[i][k] = '0';for (int z = 0; z < 4; z++) {sink(grid, i + d[z], k + d[z + 1]);}}
}

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算法：求岛屿的数量200. Number of Islands

200. Number of Islands

1. 递归解法

2. 递归求解，优化掉重复的sink函数

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