[leetcode]695. Max Area of Island小岛最大面积

Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

Find the maximum area of an island in the given 2D array. (If there is no island, the maximum area is 0.)

Example 1:

[[0,0,1,0,0,0,0,1,0,0,0,0,0],
 [0,0,0,0,0,0,0,1,1,1,0,0,0],
 [0,1,1,0,1,0,0,0,0,0,0,0,0],
 [0,1,0,0,1,1,0,0,1,0,1,0,0],
 [0,1,0,0,1,1,0,0,1,1,1,0,0],
 [0,0,0,0,0,0,0,0,0,0,1,0,0],
 [0,0,0,0,0,0,0,1,1,1,0,0,0],
 [0,0,0,0,0,0,0,1,1,0,0,0,0]]

Given the above grid, return 6. Note the answer is not 11, because the island must be connected 4-directionally.

Example 2:

[[0,0,0,0,0,0,0,0]]

Given the above grid, return 0.

Note: The length of each dimension in the given grid does not exceed 50.

Solution1： DFS

code

class Solution {
   public int maxAreaOfIsland(int[][] grid) {
        // corner case 
        int max = 0;
        if ( grid == null || grid.length==0 || grid[0].length==0) return 0;    
        for (int i = 0; i< grid.length;i++) {
            for (int j = 0; j<grid[i].length;j++) {
                if (grid[i][j]== 1) {
                    int[] count = new int[1];
                    helper(grid, i, j, count);
                    max = Math.max(max , count[0]);
                }
            }
        }
        return max;
    }
    
    private void helper(int[][] grid, int i, int j, int[] count) {
        // base case
        if ((i<0) || (j<0) || (i>=grid.length) || (j>=grid[0].length)
           || grid[i][j] != 1 ) {            
            return;
        }   
        // recursive rule 
        count[0]++;
        grid[i][j] = -1;  // mark visited
        helper(grid, i+1, j, count);
        helper(grid, i-1, j, count);
        helper(grid, i, j+1, count);
        helper(grid, i, j-1, count);
    }   
}