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695. Max Area of Island

时间:2017-12-17 10:27:19      阅读:218      评论:0      收藏:0      [点我收藏+]

Problem

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.

Intuition & Algorithm
  Traverse. When we have visited one position, we change the character to `-1`.
  when we at `(i, j)`, check `(i, j+1), (i, j-1), (i-1, j), (i+1, j)`, if `grid[a][b] = 1, area ++`

Code

class Solution {
    public int maxAreaOfIsland(int[][] grid) {
        int n = grid.length;
        int m = grid[0].length;
        int res = 0;
        for(int i = 0; i<n;i++){
            for(int j = 0; j<m; j++){
                int curArea = 0;
                if(grid[i][j]==1){
                    curArea = findMoreIsland(grid, i, j);
                }
                res = Math.max(res, curArea);
            }
        }
        return res;
    }

    private int findMoreIsland(int[][] grid, int i, int j){
        if(i<0 || j>=grid[0].length || i>=grid.length || j<0 || grid[i][j] != 1) return 0;
        grid[i][j] = -1;
        int plus = 1;
        plus += findMoreIsland(grid, i, j-1) + findMoreIsland(grid, i, j+1)
                + findMoreIsland(grid, i-1, j) + findMoreIsland(grid, i+1, j);
        return plus;
    }
}

 

695. Max Area of Island

原文:http://www.cnblogs.com/zhengqw-77/p/8051440.html

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