题目
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
1. m == grid.length
2. n == grid[i].length
3. 1 <= m, n <= 300
4. grid[i][j] 的值为 '0' 或 '1'
思路
因为此题的代码和思路都与力扣419甲板上的战舰这题非常的相似，所以今天直接来分析这两题
首先题目给出的依旧是一个二维网格，每个点值为**'1’代表陆地**，值为**'0’代表水域**，并且陆地之前相连的方向也规定为水平或竖直，所以这题我们只需要对这个二维网格进行遍历，然后在遍历的点值为’1’的时候进入dfs，在dfs中，我们标记当前搜索的点，并且向上下左右四个方向进行搜索，最后在记录进入dfs的次数就是岛屿的数量了，下面看代码
代码
首先是主函数，进行一次遍历并且条件是当前点值为1时进入dfs函数，同时记录次数
class Solution {
public:
int m, n;
int count = 0;
int numIslands(vector<vector<char>>& grid) {
m = int(grid.size());
n = int(grid[0].size());
for(int i = 0; i < m; i++) {
for(int j = 0; j < n; j++) {
if(grid[i][j] == '1') {
dfs(grid, i, j);
++count;
}
}
}
return count;
}
};
然后在dfs中，出口条件还是坐标越界或者当前搜索的点值不等于’1’或等于我们做标记的值
void dfs(vector<vector<char>>& grid, int x, int y) {
if(x < 0 || x >= m || y < 0 || y >= n || grid[x][y] !='1' || grid[x][y] == 'Q') {
return;
}
}
出口条件写完后，我们就需要对当前点进行标记，然后对当前点的上下左右进行访问
void dfs(vector<vector<char>>& grid, int x, int y) {
if(x < 0 || x >= m || y < 0 || y >= n || grid[x][y] !='1' || grid[x][y] == 'Q') {
return;
}
grid[x][y] = 'Q';
dfs(grid, x + 1, y);
dfs(grid, x - 1, y);
dfs(grid, x, y + 1);
dfs(grid, x, y - 1);
}
最后把代码整合起来
完整代码：
class Solution {
public:
int m, n;
void dfs(vector<vector<char>>& grid, int x, int y) {
if(x < 0 || x >= m || y < 0 || y >= n || grid[x][y] !='1' || grid[x][y] == 'Q') {
return;
}
grid[x][y] = 'Q';
dfs(grid, x + 1, y);
dfs(grid, x - 1, y);
dfs(grid, x, y + 1);
dfs(grid, x, y - 1);
}
int numIslands(vector<vector<char>>& grid) {
m = int(grid.size());
n = int(grid[0].size());
int count = 0;
for(int i = 0; i < m; i++) {
for(int j = 0; j < n; j++) {
if(grid[i][j] == '1') {
dfs(grid, i, j);
++count;
}
}
}
return count;
}
};
然后是419题，因为那题也是求战舰的数量，其实质与这一题是完全一样的，下面我把完整代码放出来供大家对比参考
419题完整代码：
class Solution {
public:
int m, n;
void dfs(vector<vector<char>>& board, int x, int y) {
if(x < 0 || x >= m || y < 0 || y >= n || board[x][y] != 'X' || board[x][y] == 'A') {
return;
}
board[x][y] = 'A';
dfs(board, x + 1, y);
dfs(board, x - 1, y);
dfs(board, x, y + 1);
dfs(board, x, y - 1);
}
int countBattleships(vector<vector<char>>& board) {
m = int(board.size());
n = int(board[0].size());
int ans = 0;
for(int i = 0; i < m; i++) {
for(int j = 0; j < n; j++) {
if(board[i][j] == 'X') {
dfs(board, i, j);
++ans;
}
}
}
return ans;
}
};