LeetCode 695. Max Area of Island

转载

mb5fd33ffac6eab 2017-10-29 13:55:00

文章标签 LeetCode Array DFS 2d i++ 文章分类 代码人生

原题链接在这里：https://leetcode.com/problems/max-area-of-island/description/

题目：

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.

题解：

对grid每个为1的格子做dfs, 求该岛的面积. 维护最大面积.

Time Complexity: O(m*n). m = grid.length, n = grid[0].length.

Space: O(m*n). stack space.

AC Java:

 1 class Solution {
 2     public int maxAreaOfIsland(int[][] grid) {
 3         int res = 0;
 4         for(int i = 0; i<grid.length; i++){
 5             for(int j = 0; j<grid[0].length; j++){
 6                 if(grid[i][j] == 1){
 7                     res = Math.max(res, findMaxArea(grid, i, j));
 8                 }
 9             }
10         }
11         
12         return res;
13     }
14     
15     private int findMaxArea(int [][] grid, int i, int j){
16         int area = 0;
17         if(i>=0 && i<grid.length && j>=0 && j<grid[0].length && grid[i][j]==1){
18             grid[i][j] = 0;
19             area = 1 + findMaxArea(grid,i-1,j) + findMaxArea(grid,i+1,j) + findMaxArea(grid,i,j-1) + findMaxArea(grid,i,j+1);
20         }
21         return area;
22     }
23 }

也可bfs.

Time Complexity: O(m*n). m = grid.length. n = grid[0].length.

Space: O(m*n).

AC Java:

 1 class Solution {
 2     public int maxAreaOfIsland(int[][] grid) {
 3         if(grid == null || grid.length == 0 || grid[0].length == 0){
 4             return 0;
 5         }
 6         
 7         int m = grid.length;
 8         int n = grid[0].length;
 9         int res = 0;
10         
11         for(int i = 0; i<m; i++){
12             for(int j = 0; j<n; j++){
13                 if(grid[i][j] == 1){
14                     int area = bfs(grid, i, j);
15                     res = Math.max(res, area);
16                 }
17             }
18         }
19         return res;
20     }
21     
22     int [][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
23     
24     private int bfs(int [][] grid, int i, int j){
25         int area = 1;
26         
27         grid[i][j] = 0;
28         LinkedList<int []> que = new LinkedList<int []>();
29         que.add(new int[]{i, j});
30         while(!que.isEmpty()){
31             int [] cur = que.poll();
32             for(int [] dir : dirs){
33                 int di = cur[0] + dir[0];
34                 int dj = cur[1] + dir[1];
35                 if(di>=0 && di<grid.length && dj>=0 && dj<grid[0].length && grid[di][dj]==1){
36                     grid[di][dj] = 0;
37                     area++;
38                     que.add(new int[]{di, dj});
39                 }
40             }
41         }
42         
43         return area;
44     }
45 }