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LeetCode 面试经典150题 [81/150 岛屿数量]
2024-07-31

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

个人C++解答#

递归搜索,果然一使用递归就有点不会了,说是深度优先搜索,但是没感觉到像

class Solution {
public:
void dfs(vector<vector<char>>& grid, int i, int j) {
int i_num = grid.size();
int j_num = grid[0].size();
grid[i][j] = '0';
if (i - 1 >= 0 && grid[i - 1][j] == '1')
dfs(grid, i - 1, j);
if (i + 1 < i_num && grid[i + 1][j] == '1')
dfs(grid, i + 1, j);
if (j - 1 >= 0 && grid[i][j - 1] == '1')
dfs(grid, i, j - 1);
if (j + 1 < j_num && grid[i][j + 1] == '1')
dfs(grid, i, j + 1);
}
int numIslands(vector<vector<char>>& grid) {
int i_num = grid.size();
if (!i_num)
return 0;
int j_num = grid[0].size();
int ans = 0;
for (int i = 0; i < i_num; i++) {
for (int j = 0; j < j_num; j++) {
if (grid[i][j] == '1') {
++ans;
dfs(grid, i, j);
}
}
}
return ans;
}
};
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LeetCode 面试经典150题 [81/150 岛屿数量]
https://ericonline.cn/posts/leetcode-面试经典150题-81-150-岛屿数量/
作者
Eric
发布于
2024-07-31
许可协议
CC BY-NC-SA 4.0

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