LeetCode695. Max Area of Island

一、题目

You are given an m x n binary matrix grid. 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.

The area of an island is the number of cells with a value 1 in the island.

Return the maximum area of an island in grid. If there is no island, return 0.

Example 1:

Input: grid = [[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]]
Output: 6
Explanation: The answer is not 11, because the island must be connected 4-directionally.
Example 2:

Input: grid = [[0,0,0,0,0,0,0,0]]
Output: 0

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 50
grid[i][j] is either 0 or 1.

二、题解

class Solution {
public:
    int count = 1;
    int dirs[4][2] = {0, 1, 1, 0, -1, 0, 0, -1};
    void dfs(vector<vector<int>>& grid,vector<vector<bool>>& visted,int x,int y){
        for(int i = 0;i < 4;i++){
            int nextX = x + dirs[i][0];
            int nextY = y + dirs[i][1];
            if(nextX < 0 || nextX >= grid.size() || nextY < 0 || nextY >= grid[0].size()) continue;
            if(grid[nextX][nextY] == 1 && visted[nextX][nextY] == false){
                count++;
                visted[nextX][nextY] = true;
                dfs(grid,visted,nextX,nextY);
            }
        }
    }
    int maxAreaOfIsland(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        int res = 0;
        vector<vector<bool>> visted(m,vector<bool>(n,false));
        for(int i = 0;i < m;i++){
            for(int j = 0;j < n;j++){
                if(grid[i][j] == 1 && visted[i][j] == false){
                    count = 1;
                    visted[i][j] = true;
                    dfs(grid,visted,i,j);
                    res = max(res,count);
                }
            }
        }
        return res;
    }
};