leetcode刷题. 463. 659. 岛屿问题的bfs解决方法

本文链接：https://blog.youkuaiyun.com/paradox_1_0/article/details/105523666

本文介绍了一种计算二维网格中岛屿周长的算法，通过遍历每个单元格来确定边界；同时，还详细解释了如何使用广度优先搜索（BFS）算法找出岛屿的最大面积，包括初始化队列、标记已访问区域并计算面积等步骤。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

    int islandPerimeter(vector<vector<int>>& grid) {

        int xlen = grid.size();
        int ylen = grid[0].size();

        int perimeter = 0;

        for(int i = 0; i < xlen; i++) {

            for(int j = 0; j < ylen; j++) {

                if(grid[i][j] == 0)
                    continue;

                if(i == 0 || (i > 0 && grid[i - 1][j] == 0))
                    perimeter++;

                if(i + 1 == xlen || (i + 1 < xlen && grid[i + 1][j] == 0))
                    perimeter++;

                if(j == 0 || (j > 0 && grid[i][j - 1] == 0))
                    perimeter++;

                if(j + 1 == ylen || (j + 1 < ylen && grid[i][j + 1] == 0))
                    perimeter++;
            }
        }

        return perimeter;
    }

695. 岛屿的最大面积

    int maxAreaOfIsland(vector<vector<int>>& grid) {

        int xlen = grid.size();
        int ylen = grid[0].size();

        int maxS = 0;
        // 基于bfs的保存队列
        queue<pair<int, int>> q;

        for(int i = 0; i < xlen; i++) {

            for(int j = 0; j < ylen; j++) {

                if(grid[i][j] == 0)
                    continue;

                // 记录当前面积
                int S = 0;
                q.push(pair<int, int>(i, j));

                while(q.size()) {

                    S++;
                    pair<int, int> point = q.front();
                    q.pop();
                    grid[point.first][point.second] = 0;

                // 上
                    if(point.first > 0 && grid[point.first - 1][point.second] == 1) {

                        q.push(pair<int, int>(point.first - 1, point.second));
                        grid[point.first - 1][ point.second] = 0;
                    }
                // 下
                    if(point.first + 1 < xlen && grid[point.first + 1][point.second] == 1) {

                        q.push(pair<int, int>(point.first + 1, point.second));
                        grid[point.first + 1][point.second] = 0;
                    }
                // 左
                    if(point.second > 0 && grid[point.first][point.second - 1] == 1) {

                        q.push(pair<int, int>(point.first, point.second - 1));
                        grid[point.first][point.second - 1] = 0;
                    }
                // 右
                    if(point.second + 1 < ylen && grid[point.first][point.second + 1] == 1) {

                        q.push(pair<int, int>(point.first, point.second + 1));
                        grid[point.first][point.second + 1] = 0;
                    }
                }

                if(maxS < S) maxS = S;
            }
        }
        return maxS;
    }