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最大人工岛⚓︎

Leetcode题目链接

描述⚓︎

详见中文题目链接。

解答⚓︎

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class Solution {
private:
    int m, n;
    int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
    void dfs(vector<vector<int>>& grid, int i, int j, int id) {
        if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] != 1) return;
        grid[i][j] = id;  // if grid[i][j] == 1
        for (int k = 0; k < 4; k++) {
            dfs(grid, i + dx[k], j + dy[k], id);
        }
    }

public:
    int largestIsland(vector<vector<int>>& grid) {
        m = grid.size(); n = grid[0].size();
        int id = 2;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                dfs(grid, i, j, id++);
            }
        }
        unordered_map<int, int> sizeMap;
        int maxSize = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] > 1) {
                    maxSize = max(maxSize, ++sizeMap[grid[i][j]]);
                }
            }
        }

        vector<bool> visited(id, false);
        int up, down, left, right, merge;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 0) {
                    up = (i > 0) ? grid[i - 1][j] : 0;
                    down = (i + 1 < m) ? grid[i + 1][j] : 0;
                    left = (j > 0) ? grid[i][j - 1] : 0;
                    right = (j + 1 < n) ? grid[i][j + 1] : 0;

                    visited[up] = true;
                    merge = 1 + sizeMap[up];
                    if (!visited[down]) {
                        merge += sizeMap[down]; visited[down] = true;
                    }
                    if (!visited[left]) {
                        merge += sizeMap[left]; visited[left] = true;
                    }
                    if (!visited[right]) {
                        merge += sizeMap[right]; visited[right] = true;
                    }
                    maxSize = max(maxSize, merge);

                    visited[up] = false;
                    visited[down] = false;
                    visited[left] = false;
                    visited[right] = false;
                }
            }
        }
        return maxSize;
    }
};