示例 1:
输入:grid = [[1,1],[1,2]], r0 = 0, c0 = 0, color = 3输出:[[3, 3], [3, 2]]示例 2:
输入:grid = [[1,2,2],[2,3,2]], r0 = 0, c0 = 1, color = 3输出:[[1, 3, 3], [2, 3, 3]]示例 3:
输入:grid = [[1,1,1],[1,1,1],[1,1,1]], r0 = 1, c0 = 1, color = 2输出:[[2, 2, 2], [2, 1, 2], [2, 2, 2]] 思路 这道题目可以使用递归来做但是要注意只有边界才会着色, 也就是说 1 1 11 1 11 1 1112这样的测试用例最中间的1不是边界, 所以这个1不用着色之 这样的边界条件就是四周都是和当前的位置一样就不用着色 代码 class Solution {public: void dfs(vector<vector<int>>& res, vector<vector<int>>& grid, vector<vector<bool>>& vis, int r, int c, int color, int oldcolor) { if (r < 0 || r >= grid.size() || c < 0 || c >= grid[0].size() || grid[r][c] != oldcolor || vis[r][c] == true) return; if (r > 0 && r < grid.size() - 1 && c > 0 && c < grid[0].size() - 1 && grid[r - 1][c] == oldcolor && grid[r + 1][c] == oldcolor && grid[r][c + 1] == oldcolor && grid[r][c - 1] == oldcolor) { ; } else { res[r][c] = color; } vis[r][c] = true; dfs(res, grid, vis, r + 1, c, color, oldcolor); dfs(res, grid, vis, r - 1, c, color, oldcolor); dfs(res, grid, vis, r, c + 1, color, oldcolor); dfs(res, grid, vis, r, c - 1, color, oldcolor); } vector<vector<int>> colorBorder(vector<vector<int>>& grid, int r0, int c0, int color) { vector<vector<bool>> vis(grid.size(), vector<bool>(grid[0].size(), false)); vector<vector<int>> res = grid; dfs(res, grid, vis, r0, c0, color, grid[r0][c0]); return res; }};