200. Number of Islands | Java最短代码实现

原题链接：200. Number of Islands
【思路】

本题考查dfs和bfs的结合。由于题目没有限制不可改变原数组的结构，所以一旦遇到1，就将周边相邻的几个1置为2。遍历整个数组，遇到1的个数要求的解：

public int numIslands(char[][] grid) {
int islansNum = 0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (grid[i][j] != '1') continue;
dfs(grid, i, j);
islansNum++;
}
}
return islansNum;
}
public void dfs(char[][] grid, int row, int column) {
if (row >= 0 && row < grid.length
&& column >= 0 && column < grid[0].length) {
if (grid[row][column] == '1') {
grid[row][column] = '2';
dfs(grid, row + 1, column);
dfs(grid, row, column + 1);
dfs(grid, row - 1, column);
dfs(grid, row, column - 1);
}
}
}

45 / 45 test
cases passed. Runtime: 4
ms Your runtime beats 37.40% of javasubmissions.

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