欢迎访问LeetCode题解合集

题目描述

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'

题解：

深搜。

这题就是求 联通块 的数量。可以遍历 grid 中的每个元素，对每个 1 ，从该位置进行搜索，将与它相邻的 1 做一下标记即可。

时间复杂度：$O(n * m)$

额外空间复杂度：$O(n * m)$

class Solution {
public:
    vector<int> dx, dy;
    int n, m;
    void dfs( int x, int y, vector<vector<char>>& grid ){
        grid[x][y] = '0';
        for( int k = 0; k < 4; ++k ) {
            int tx = x + dx[k];
            int ty = y + dy[k];
            if( tx < 0 || tx >= n || ty < 0 || ty >= m || grid[tx][ty] == '0' )
                continue;
            dfs( tx, ty, grid );
        }
    }
    int numIslands(vector<vector<char>>& grid) {
        dx = { -1, 0, 1, 0 };
        dy = { 0, 1, 0, -1 };
        n = grid.size();
        m = grid[0].size();
        int num = 0;
        for ( int i = 0; i < n; ++i ) {
            for ( int j = 0; j < m; ++j ) {
                if( grid[i][j] == '0' ) continue;
                ++num;
                dfs( i, j, grid );
            }
        }
        return num;
    }
};
/*
时间：16ms，击败：91.16%
内存：9.3MB，击败：81.93%
*/