/**
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'.
Related Topics深度优先搜索 | 广度优先搜索 | 并查集 | 数组 | 矩阵
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*/
//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
public int numIslands(char[][] grid) {
int cnt = 0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (grid[i][j] == '1') {
cnt++;
scaleOut(grid, i, j);
}
}
}
return cnt;
}
private void scaleOut(char[][] grid, int r, int c) {
if (grid[r][c] == '0') {
return;
}
grid[r][c] = '0';
if (r < grid.length - 1) {
scaleOut(grid, r + 1, c);
}
if (c < grid[0].length - 1) {
scaleOut(grid, r, c + 1);
}
if (r > 0) {
scaleOut(grid, r - 1, c);
}
if (c > 0) {
scaleOut(grid, r, c - 1);
}
}
}
//leetcode submit region end(Prohibit modification and deletion)
