Question
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’
Related Topics 深度优先搜索 广度优先搜索 并查集 数组 矩阵 👍 1733 👎 0
Answer
//200
//11110
//11010
//11000
//00000
//1
//200-Solution1
void bfs(int i, int j, int n, int m, vector<vector<char>>& grid) {
if (i<0 || i>n-1 || j<0 || j>m-1 || grid[i][j]=='0')
//[0~n-1],递归边界是出界或者找到为0的值
return;
grid[i][j]='0';//搜索到为1的值置0，不用再次遍历
bfs(i-1, j, n, m, grid);
bfs(i+1, j, n, m, grid);
bfs(i, j-1, n, m, grid);
bfs(i, j+1, n, m, grid);//上下左右，可以换顺序
}
int numIslands0(vector<vector<char>>& grid) {
if (grid.empty()) {
return 0;
}
int n = grid.size();
int m = grid[0].size();
int res = 0;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
if (grid[i][j] == '1') {
bfs(i, j, n, m, grid);
res++;
}
}
}
return res;
}
//200-Solution2
void dfs(int i, int j, int n, int m, vector<vector<char>>& grid) {
grid[i][j] = 0;
if (i>0 && grid[i-1][j] == '1') {
dfs(i-1, j, n, m, grid);
}
if (i<n-1 && grid[i+1][j] == '1') {
dfs(i+1, j, n, m, grid);
}
if (j>0 && grid[i][j-1] == '1') {
dfs(i, j-1, n, m, grid);
}
if (j<m-1 && grid[i][j+1] == '1') {
dfs(i, j+1, n, m, grid);
}
}
int numIslands(vector<vector<char>>& grid) {
if (grid.empty()) {
return 0;
}
int n = grid.size();
int m = grid[0].size();
int res = 0;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
if (grid[i][j] == '1') {
dfs(i, j, n, m, grid);
res++;
}
}
}
return res;
}
