3.15 岛屿的最大面积

题目

给定一个包含了一些 0 和 1的非空二维数组 grid , 一个 岛屿 是由四个方向 (水平或垂直) 的 1 (代表土地) 构成的组合。你可以假设二维矩阵的四个边缘都被水包围着。

找到给定的二维数组中最大的岛屿面积。(如果没有岛屿，则返回面积为0。)

示例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]]

思路

• 遍历每一个坐标作为开始，每一个坐标都去递归的遍历上下左右，遍历过的标记为遍历过，下次就不再遍历。记录每一个坐标作为开始时的岛屿面积，取最大的。

代码

class Solution {
private:
    vector<vector<bool>> used;
    vector<vector<int>> pattern;
    int rows;
    int cols;
    int maxArea;
public:
    int maxAreaOfIsland(vector<vector<int>>& grid) {
        if ( grid.size() == 0 ) return 0;
        
        rows = grid.size();
        cols = grid[0].size();
        maxArea = 0;

        used = vector<vector<bool>>( rows, vector<bool>( cols, false ) );
        pattern = { {1, 0}, {-1, 0}, {0, 1}, {0, -1} };

        for ( int i = 0; i < rows; ++i ) {
            for ( int j = 0; j < cols; ++j ) {
                if ( !used[i][j] && grid[i][j] == 1 ) {
                    int temp = 1;
                    helper( grid, i, j, temp );
                    maxArea = max( maxArea, temp );
                }
            }
        }

        return maxArea;
    }

    void helper( vector<vector<int>>& grid, int row, int col, int& temp ) {
        used[row][col] = true;
        
        for ( auto& vec : pattern ) {
            int newRow = row + vec[0];
            int newCol = col + vec[1];

            if ( isValid(grid, newRow, newCol) && !used[newRow][newCol] ) {
                ++temp;

                helper( grid, newRow, newCol, temp );
            }
        }

        return;
    }

    bool isValid(vector<vector<int>>& grid, int& row, int& col ) {
        return row >=0 && row < rows 
            && col >= 0 && col < cols
            && grid[row][col] == 1;
    }
};