leetcode 695. Max Area of Island

最新推荐文章于 2024-07-29 23:53:47 发布

转载最新推荐文章于 2024-07-29 23:53:47 发布 · 216 阅读

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原文链接：http://www.cnblogs.com/qinduanyinghua/p/11551112.html

本文介绍了一种使用深度优先搜索(DFS)算法来计算二维数组中最大岛屿面积的方法。通过遍历数组，对每个岛屿进行DFS搜索，计算其面积，并更新最大面积。适用于岛屿由1表示且四向连接的情况。

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.

思路：计算每个岛屿的面积。（深搜）

 1 class Solution {
 2 public:
 3     int maxAreaOfIsland(vector<vector<int>>& grid) {
 4         int maxArea = 0;
 5         int m = grid.size();
 6         if (m == 0)
 7             return 0;
 8         int n = grid[0].size();
 9         for (int i = 0; i < m; i++) {
10             for (int j = 0; j < n; j++) {
11                 if (grid[i][j] == 1) {
12                     int tmp = dfs(i, j, grid);
13                     maxArea = max(maxArea, tmp);
14                 }
15             }
16         }
17         return maxArea;
18     }
19 private:
20     int dfs(int i, int j, vector<vector<int> > &grid) {
21         grid[i][j] = 0;
22         int cnt = 1;
23         int dx[4] = {-1, 0, 1, 0};
24         int dy[4] = {0, 1, 0, -1};
25         //four dimension
26         for (int d = 0; d < 4; d++) {
27             int newx = i + dx[d];
28             int newy = j + dy[d];
29             if (newx >= 0 && newx < grid.size() && newy >= 0 && newy < grid[0].size() && grid[newx][newy] == 1) {
30                 cnt += dfs(newx, newy, grid);
31             }
32         }
33         return cnt;
34     }
35 };

转载于:https://www.cnblogs.com/qinduanyinghua/p/11551112.html