LeetCode/200. 岛屿数量

200. 岛屿数量

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

提示：

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 300
• grid[i][j] 的值为 '0' 或 '1'

来源：力扣（LeetCode）
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题解：

本题深度优先搜索遍历是首选，我们通过遍历二维数组，找到当前位置为一的元素，在该位置进行DFS，将当前元素置为'0'，然后向该元素的上下左右进行遍历。本题当然也可以用广度优先搜索和并查集，但是相对较为繁琐，此处不再给出详细代码。

具体代码如下：

class Solution {
    public int numIslands(char[][] grid) {
        if (grid == null || grid.length == 0) {
            return 0;
        }
        int result = 0, row = grid.length, column = grid[0].length;
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < column; j++) {
                if (grid[i][j] == '1') {
                    result++;
                    dfs(grid, i, j);
                }
            }
        }
        return result;
    }

    private void dfs(char[][] grid, int i, int j) {
        if (i < 0 || j < 0 || i >= grid.length || j >= grid[0].length || grid[i][j] == '0')
            return;

        grid[i][j] = '0';
        dfs(grid, i - 1, j);
        dfs(grid, i + 1, j);
        dfs(grid, i, j - 1);
        dfs(grid, i, j + 1);
    }
}

并查集框架：

private void union(int x, int y) {
    int xParent = find(x);
    int yParent = find(y);
    if (xParent == yParent)
        return;

    parent[x] = y;
}

private int find(int x) {
    while (x != parent[x]) {
        parent[x] = parent[parent[x]];
        x = parent[x];
    }
    return x;
}