Given a non-empty 2D array
grid
of 0's and1'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.)
[[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.[[0,0,0,0,0,0,0,0]]
Given the above grid, return
0
.
- The length of each dimension in the given
grid
does not exceed 50.
- mine
DFS Runtime: 2 ms, faster than 99.42%, Memory Usage: 39.5 MB, less than 96.30% of Java online submissions
A variant of this question is 959. Regions Cut By Slashes.
//O(R*C)Time O(1)Space
public int maxAreaOfIsland(int[][] grid) {
int res = 0;
for(int i = 0; i < grid.length; i++){
for(int j = 0; j < grid[i].length; j++){
if(grid[i][j] == 1){
res = Math.max(res, dfs(grid, i, j));
}
}
}
return res;
}
public int dfs(int[][] grid,int i,int j){
if(i>=0 && i< grid.length && j >= 0 && j < grid[i].length && grid[i][j] == 1){
grid[i][j] = 0;
return 1 + dfs(grid, i, j-1) + dfs(grid, i, j+1)
+ dfs(grid, i - 1, j) + dfs(grid, i + 1, j);
}
return 0;
}
- the most votes
Your runtime beats 87.82 % of java submissions.
class Solution {
public int maxAreaOfIsland(int[][] grid) {
int res = 0;
int row = grid.length;
int column = grid[0].length;
for(int i = 0; i < row; i++){
for(int j = 0; j < column; j++){
if(grid[i][j] == 1){
res = Math.max(res,AreaOfIsland(grid,i,j));
}
}
}
return res;
}
public int AreaOfIsland(int[][] grid, int i, int j){
if( i >= 0 && i < grid.length && j >= 0 && j < grid[0].length && grid[i][j] == 1){
grid[i][j] = 0;
return 1 + AreaOfIsland(grid, i+1, j) + AreaOfIsland(grid, i-1, j) + AreaOfIsland(grid, i, j-1) + AreaOfIsland(grid, i, j+1);
}
return 0;
}
}