README.md

1034. 边界着色

English Version

题目描述

给你一个大小为 m x n 的整数矩阵 grid ，表示一个网格。另给你三个整数 row、col 和 color 。网格中的每个值表示该位置处的网格块的颜色。

两个网格块属于同一 连通分量 需满足下述全部条件：

• 两个网格块颜色相同
• 在上、下、左、右任意一个方向上相邻

连通分量的边界 是指连通分量中满足下述条件之一的所有网格块：

• 在上、下、左、右任意一个方向上与不属于同一连通分量的网格块相邻
• 在网格的边界上（第一行/列或最后一行/列）

请你使用指定颜色 color 为所有包含网格块 grid[row][col] 的 连通分量的边界 进行着色，并返回最终的网格 grid 。

 

示例 1：

输入：grid = [[1,1],[1,2]], row = 0, col = 0, color = 3
输出：[[3,3],[3,2]]

示例 2：

输入：grid = [[1,2,2],[2,3,2]], row = 0, col = 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]], row = 1, col = 1, color = 2
输出：[[2,2,2],[2,1,2],[2,2,2]]

 

提示：

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 50
• 1 <= grid[i][j], color <= 1000
• 0 <= row < m
• 0 <= col < n

 

解法

深度优先搜索，利用 vis 记录访问过的位置。

Python3

class Solution:
    def colorBorder(
        self, grid: List[List[int]], row: int, col: int, color: int
    ) -> List[List[int]]:
        m, n = len(grid), len(grid[0])
        vis = [[False] * n for _ in range(m)]

        def dfs(i, j, color):
            vis[i][j] = True
            old_color = grid[i][j]
            for a, b in [[-1, 0], [1, 0], [0, -1], [0, 1]]:
                x, y = a + i, b + j
                if 0 <= x < m and 0 <= y < n:
                    if not vis[x][y]:
                        if grid[x][y] == old_color:
                            dfs(x, y, color)
                        else:
                            grid[i][j] = color
                else:
                    grid[i][j] = color

        dfs(row, col, color)
        return grid

Java

class Solution {
    private int[] dirs = new int[]{-1, 0, 1, 0, -1};

    public int[][] colorBorder(int[][] grid, int r0, int c0, int color) {
        boolean[][] vis = new boolean[grid.length][grid[0].length];
        dfs(grid, r0, c0, color, vis);
        return grid;
    }

    private void dfs(int[][] grid, int i, int j, int color, boolean[][] vis) {
        vis[i][j] = true;
        int oldColor = grid[i][j];
        for (int k = 0; k < 4; ++k) {
            int x = i + dirs[k], y = j + dirs[k + 1];
            if (x >= 0 && x < grid.length && y >= 0 && y < grid[0].length) {
                if (!vis[x][y]) {
                    if (grid[x][y] == oldColor) {
                        dfs(grid, x, y, color, vis);
                    } else {
                        grid[i][j] = color;
                    }
                }
            } else {
                grid[i][j] = color;
            }
        }
    }
}

C++

class Solution {
public:
    int m, n;
    vector<vector<int>> dirs = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

    vector<vector<int>> colorBorder(vector<vector<int>>& grid, int row, int col, int color) {
        m = grid.size();
        n = grid[0].size();
        vector<vector<bool>> vis(m, vector<bool>(n, false));
        dfs(row, col, color, grid, vis);
        return grid;
    }

    void dfs(int i, int j, int color, vector<vector<int>>& grid, vector<vector<bool>>& vis) {
        vis[i][j] = true;
        int oldColor = grid[i][j];
        for (auto& dir : dirs) {
            int x = i + dir[0], y = j + dir[1];
            if (x >= 0 && x < m && y >= 0 && y < n) {
                if (!vis[x][y]) {
                    if (grid[x][y] == oldColor)
                        dfs(x, y, color, grid, vis);
                    else
                        grid[i][j] = color;
                }
            } else
                grid[i][j] = color;
        }
    }
};

Go

func colorBorder(grid [][]int, row int, col int, color int) [][]int {
	m, n := len(grid), len(grid[0])
	vis := make([][]bool, m)
	for i := 0; i < m; i++ {
		vis[i] = make([]bool, n)
	}
	dirs := [4][2]int{{0, -1}, {0, 1}, {1, 0}, {-1, 0}}

	var dfs func(i, j, color int)
	dfs = func(i, j, color int) {
		vis[i][j] = true
		oldColor := grid[i][j]
		for _, dir := range dirs {
			x, y := i+dir[0], j+dir[1]
			if x >= 0 && x < m && y >= 0 && y < n {
				if !vis[x][y] {
					if grid[x][y] == oldColor {
						dfs(x, y, color)
					} else {
						grid[i][j] = color
					}
				}
			} else {
				grid[i][j] = color
			}
		}
	}
	dfs(row, col, color)
	return grid
}

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