[LeetCode] 695. Max Area of Island

轉自LeetCode

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.

<Solution>

這題是典型的DFS，用DFS的觀念解即可

code 如下

C++

	class Solution {
	public:
	int maxAreaOfIsland(vector<vector<int>>& grid) {
	const int row = grid.size();
	const int col = grid[0].size();
	int ans = INT_MIN;
	int cnt;
	queue<pair<int,int>> dfs;
	pair<int,int> land;
	for(int r = 0; r < row; ++r) {
	for(int c = 0; c < col; ++c) {
	if(grid[r][c] == 1) {
	cnt = 0;
	grid[r][c] = -1;
	dfs.push(make_pair(r,c));
	while(!dfs.empty()) {
	land = dfs.front();
	dfs.pop();
	++cnt;
	//>> up
	if(land.first > 0 && grid[land.first-1][land.second] == 1) {
	grid[land.first-1][land.second] = -1;
	dfs.push(make_pair(land.first-1,land.second));
	}
	//>> down
	if(land.first+1 < row && grid[land.first+1][land.second] == 1) {
	grid[land.first+1][land.second] = - 1;
	dfs.push(make_pair(land.first+1,land.second));
	}
	//>> left
	if(land.second > 0 && grid[land.first][land.second-1] == 1) {
	grid[land.first][land.second-1] = -1;
	dfs.push(make_pair(land.first,land.second-1));
	}
	//>> right
	if(land.second+1 < col && grid[land.first][land.second+1] == 1) {
	grid[land.first][land.second+1] = -1;
	dfs.push(make_pair(land.first,land.second+1));
	}
	}
	ans = max(ans, cnt);
	}
	}
	}
	return ans == INT_MIN ? 0 : ans;
	}
	};

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Java

	class Solution {
	public int maxAreaOfIsland(int[][] grid) {
	int ans = 0;
	int cnt;
	Stack<int[]> dfs = new Stack<>();
	int[] tmp;
	for(int r = 0; r < grid.length; ++r) {
	for(int c = 0; c < grid[0].length; ++c) {
	if(grid[r][c] == 1) {
	cnt = 0;
	grid[r][c] = -1;
	dfs.push(new int[]{r, c});
	while(!dfs.empty()) {
	++cnt;
	tmp = dfs.pop();
	//>> up
	if(tmp[0] > 0 && grid[tmp[0]-1][tmp[1]] == 1) {
	grid[tmp[0]-1][tmp[1]] = -1;
	dfs.push(new int[]{tmp[0]-1, tmp[1]});
	}
	//>> down
	if(tmp[0] + 1 < grid.length && grid[tmp[0]+1][tmp[1]] == 1) {
	grid[tmp[0]+1][tmp[1]] = -1;
	dfs.push(new int[]{tmp[0]+1, tmp[1]});
	}
	//>> left
	if(tmp[1] > 0 && grid[tmp[0]][tmp[1]-1] == 1) {
	grid[tmp[0]][tmp[1]-1] = -1;
	bfs.push(new int[]{tmp[0], tmp[1]-1});
	}
	//>> right
	if(tmp[1]+1 < grid[0].length && grid[tmp[0]][tmp[1]+1] == 1) {
	grid[tmp[0]][tmp[1]+1] = -1;
	bfs.push(new int[]{tmp[0], tmp[1]+1});
	}
	}
	ans = Math.max(ans, cnt);
	}
	}
	}
	return ans;
	}
	}

view raw maxAreaOfIsland.java hosted with ❤ by GitHub