个人练习-Leetcode-1034. Coloring A Border

题目链接https://leetcode.cn/problems/coloring-a-border/

题目大意：给出一个二维矩阵grid，元素是int代表颜色。给出一个坐标row, col和一种颜色color，将这个坐标的连通分量的边界染成color的颜色。返回这个新的grid

思路：先DFS找连通分量，再遍历一遍将非边界的去掉就得到边界，再染色即可。当然写起来还是要注意一些小细节的。

一个二维vectorkNown[]保存是否访问过，用于DFS；
一个二维vectorborder[]保存是否为边界；

一个注意点：如果某个格子就在grid的边界上且属于连通分量，那它一定属于连通分量的边界，因为连通分量不能再向外扩展了。

DFS过程

 void DFS(vector<vector<int>>& grid, vector<vector<bool>>& kNown, vector<vector<bool>>& border, int origin, int row, int col) {
        if (row < 0 || row >= grid.size() || col < 0 || col >= grid[0].size())
            return;
        if (kNown[row][col]) return;
        kNown[row][col] = true;
        if (grid[row][col] == origin) {
            border[row][col] = true;
            DFS(grid, kNown, border, origin, row, col+1);
            DFS(grid, kNown, border, origin, row+1, col);
            DFS(grid, kNown, border, origin, row, col-1);
            DFS(grid, kNown, border, origin, row-1, col);
        }
    }

此时border[]内为true的是连通分量，接下来要去掉非边界的部分，遍历即可，四个方向都是相同颜色的话，它就是非边界。

        for (int i = 1; i < m-1; i++) {
            for (int j = 1; j < n-1; j++) {
                if (border[i][j]) {
                    if (grid[i+1][j] == origin && grid[i][j+1] == origin
                    && grid[i-1][j] == origin && grid[i][j-1] == origin) {
                        border[i][j] = false;
                    }
                }
            }
        }

此时border内就只剩连通分量的边界了，遍历染色即可

完整代码

class Solution {
public:
    void DFS(vector<vector<int>>& grid, vector<vector<bool>>& kNown, vector<vector<bool>>& border, int origin, int row, int col) {
        if (row < 0 || row >= grid.size() || col < 0 || col >= grid[0].size())
            return;
        if (kNown[row][col]) return;
        kNown[row][col] = true;
        if (grid[row][col] == origin) {
            border[row][col] = true;
            DFS(grid, kNown, border, origin, row, col+1);
            DFS(grid, kNown, border, origin, row+1, col);
            DFS(grid, kNown, border, origin, row, col-1);
            DFS(grid, kNown, border, origin, row-1, col);
        }
    }


    vector<vector<int>> colorBorder(vector<vector<int>>& grid, int row, int col, int color) {
        int origin = grid[row][col];
        int m = grid.size(), n = grid[0].size();
        vector<vector<bool>> kNown(m, vector<bool>(n, false));
        vector<vector<bool>> border(m, vector<bool>(n, false));

        DFS(grid, kNown, border, origin, row, col);

        for (int i = 1; i < m-1; i++) {
            for (int j = 1; j < n-1; j++) {
                if (border[i][j]) {
                    if (grid[i+1][j] == origin && grid[i][j+1] == origin
                    && grid[i-1][j] == origin && grid[i][j-1] == origin) {
                        border[i][j] = false;
                    }
                }
            }
        }

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (border[i][j]) 
                    grid[i][j] = color;              
            }
        }

        return grid;
    }
};

原文地址：https://www.jb51.cc/wenti/3281961.html