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.
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这个就是求最大岛屿的面积，1是陆地，0是海洋。这个用DFS会简单些。
C++代码：
class Solution { public: vector<vector<int>> grid; //注意dfs()里面的参数还需要添加vector<vector<int> > &grid,来保证能够dfs里面的grid的变化后，maxAreaOfIsland()里面的grid也能跟着变化。 int dfs(vector<vector<int> > &grid,int r, int c){ if(r >= 0 && r < grid.size() && c >= 0 && c < grid[0].size() && grid[r][c] == 1){ grid[r][c] = 0; return 1 + dfs(grid,r,c-1) + dfs(grid,r,c+1) + dfs(grid,r-1,c) + dfs(grid,r+1,c); } return 0; } int maxAreaOfIsland(vector<vector<int>>& grid) { int m = grid.size(); int n = grid[0].size(); if(m == 0 || n == 0) return 0; int ans = 0; for(int i = 0; i < m; i++){ for(int j = 0; j < n; j++){ if(grid[i][j] != 0) ans = max(ans,dfs(grid,i,j)); } } return ans; } };
也可以用BFS(这个是有点套路，多做几次就会熟练了)
C++代码：
int dx[] = {0,1,0,-1}; int dy[] = {1,0,-1,0}; class Solution { public: typedef pair<int,int> pii; queue<pii> q; int maxAreaOfIsland(vector<vector<int>>& grid) { int m = grid.size(); int n = grid[0].size(); if(m == 0 || n == 0) return 0; int res = 0; for(int i = 0; i < m; i++){ for(int j = 0; j < n; j++){ if(grid[i][j] == 0) continue; int cnt = 0; q.push(make_pair(i,j)); grid[i][j] = 0; while(!q.empty()){ auto t = q.front();q.pop(); cnt++; for(int i = 0; i < 4; i++){ int xx = t.first + dx[i]; int yy = t.second +dy[i]; if(xx >= 0 && xx < m && yy >= 0 && yy < n && grid[xx][yy] == 1){ grid[xx][yy] = 0; q.push(make_pair(xx,yy)); } } } res = max(res,cnt); } } return res; } };
转载于:https://www.cnblogs.com/Weixu-Liu/p/10867417.html
