827. Making A Large Island

In a 2D grid of 0s and 1s, we change at most one 0 to a 1.

After, what is the size of the largest island? (An island is a 4-directionally connected group of 1s).

Example 1:

Input: [[1, 0], [0, 1]]
Output: 3
Explanation: Change one 0 to 1 and connect two 1s, then we get an island with area = 3.

Example 2:

Input: [[1, 1], [1, 0]]
Output: 4
Explanation: Change the 0 to 1 and make the island bigger, only one island with area = 4.

Example 3:

Input: [[1, 1], [1, 1]]
Output: 4
Explanation: Can’t change any 0 to 1, only one island with area = 4.

Notes:

1 <= grid.length = grid[0].length <= 50.
0 <= grid[i][j] <= 1.

DFS求解，将数组矩阵中，值为0的地方替换为1，然后求1相邻的个数即可。直接求解的方式如下所示：

class Solution {
    
    public int largestCnt = 1, curLargestCnt = 0;
    
    public int largestIsland(int[][] grid) {
        int cnt = 0;
        boolean tag = false;
        for (int i = 0; i < grid.length; ++ i){
            for (int j = 0; j < grid[i].length; ++ j){
                curLargestCnt = 0;
                if (grid[i][j] == 0){
                    grid[i][j] = 1;
                    boolean[][] visited = new boolean[grid.length][grid[i].length];
                    largestIslandDFS(grid, visited, i, j);
                    largestCnt = Math.max(largestCnt, curLargestCnt);
                    grid[i][j] = 0;
                }
                else if (!tag){
                    tag = true;
                    boolean[][] visited = new boolean[grid.length][grid[i].length];
                    largestIslandDFS(grid, visited, i, j);
                    largestCnt = Math.max(largestCnt, curLargestCnt);
                }
            }
        }
        return largestCnt;
    }
    
    public void largestIslandDFS(int[][] grid, boolean[][] visited, int i, int j){
        if (i < 0 || i >= grid.length || j < 0 || j >= grid[i].length){
            return;
        }
        if (grid[i][j] != 1 || visited[i][j]){
            return;
        }
        visited[i][j] = true;
        curLargestCnt ++;
        largestIslandDFS(grid, visited, i, j - 1);
        largestIslandDFS(grid, visited, i - 1, j);
        largestIslandDFS(grid, visited, i + 1, j);		
        largestIslandDFS(grid, visited, i, j + 1);
    }
}

上述方案有优化的地方，就是将已经访问过的1的位置，替换为与该1相连的1的总数，这样下次访问的时候，直接取值，可以减少原始位置值为1的访问次数，提高效率。