/**
<p>給你一個由 <code>'1'</code>(陸地)和 <code>'0'</code>(水)組成的的二維網格,請你計算網格中島嶼的數量。</p>
<p>島嶼總是被水包圍,並且每座島嶼只能由水平方向和/或豎直方向上相鄰的陸地連接形成。</p>
<p>此外,你可以假設該網格的四條邊均被水包圍。</p>
<p> </p>
<p><strong>示例 1:</strong></p>
<pre>
<strong>輸入:</strong>grid = [
["1","1","1","1","0"],
["1","1","0","1","0"],
["1","1","0","0","0"],
["0","0","0","0","0"]
]
<strong>輸出:</strong>1
</pre>
<p><strong>示例 2:</strong></p>
<pre>
<strong>輸入:</strong>grid = [
["1","1","0","0","0"],
["1","1","0","0","0"],
["0","0","1","0","0"],
["0","0","0","1","1"]
]
<strong>輸出:</strong>3
</pre>
<p> </p>
<p><strong>提示:</strong></p>
<ul>
<li><code>m == grid.length</code></li>
<li><code>n == grid[i].length</code></li>
<li><code>1 <= m, n <= 300</code></li>
<li><code>grid[i][j]</code> 的值爲 <code>'0'</code> 或 <code>'1'</code></li>
</ul>
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*/
//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
//主函數計算島嶼數量
public int numIslands(char[][] grid) {
int res = 0;
int m = grid.length;
int n = grid[0].length;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if(grid[i][j]=='1'){
res++;
dfs(grid,i,j);
}
}
}
return res;
}
//子函數負責淹沒 flood fill,
void dfs(char[][] grid,int i ,int j){
int m = grid.length;
int n = grid[0].length;
if (i < 0 || j < 0 || i >= m || j >= n) {
return ;
}
if(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);
}
}
//leetcode submit region end(Prohibit modification and deletion)
leetcode200-dfs-島嶼的數量
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