add: leetcode 1034 readme

leetcode/1034.Coloring-A-Border/README.md

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+# [1034. Coloring A Border](https://leetcode-cn.com/problems/coloring-a-border/)
+
+## 题目
+
+You are given an m x n integer matrix grid, and three integers row, col, and color. Each value in the grid represents the color of the grid square at that location.
+
+Two squares belong to the same connected component if they have the same color and are next to each other in any of the 4 directions.
+
+The border of a connected component is all the squares in the connected component that are either 4-directionally adjacent to a square not in the component, or on the boundary of the grid (the first or last row or column).
+
+You should color the border of the connected component that contains the square grid[row][col] with color.
+
+Return the final grid.
+
+**Example 1**:
+
+    Input: grid = [[1,1],[1,2]], row = 0, col = 0, color = 3
+    Output: [[3,3],[3,2]]
+
+**Example 2**:
+
+    Input: grid = [[1,2,2],[2,3,2]], row = 0, col = 1, color = 3
+    Output: [[1,3,3],[2,3,3]]
+
+**Example 3**:
+
+    Input: grid = [[1,1,1],[1,1,1],[1,1,1]], row = 1, col = 1, color = 2
+    Output: [[2,2,2],[2,1,2],[2,2,2]]
+
+**Constraints:**
+
+- m == grid.length
+- n == grid[i].length
+- 1 <= m, n <= 50
+- 1 <= grid[i][j], color <= 1000
+- 0 <= row < m
+- 0 <= col < n
+
+## 题目大意
+
+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。
+
+当两个网格块的颜色相同，而且在四个方向中任意一个方向上相邻时，它们属于同一连通分量
+
+边界：在连通分量的块中（前提）并且满足以下条件之一：
+（1）要么上下左右存在一个块不在连通分量里面
+（2）要么这个块的位置在整个grid的边框上
+
+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 。
+
+## 解题思路
+
+- 用bfs进行遍历选出边界，使用color给边界着色
+
+## 代码
+
+```go
+package leetcode
+
+type point struct {
+	x int
+	y int
+}
+
+var (
+	borders       []point
+	m             int
+	n             int
+	vis           [][]bool
+	dirs          []point
+	q             []point
+	originalColor int
+)
+
+func colorBorder(grid [][]int, row, col, color int) [][]int {
+	m, n = len(grid), len(grid[0])
+	vis = make([][]bool, m)
+	for i := range vis {
+		vis[i] = make([]bool, n)
+	}
+	dirs = []point{{1, 0}, {-1, 0}, {0, 1}, {0, -1}}
+	originalColor = grid[row][col]
+	borders = []point{}
+	q = []point{{row, col}}
+	bfs(q, grid)
+	for _, p := range borders {
+		grid[p.x][p.y] = color
+	}
+	return grid
+}
+
+func bfs(q []point, grid [][]int) {
+	for len(q) != 0 {
+		ele := q[0]
+		vis[ele.x][ele.y] = true
+		q = q[1:]
+		isBorder := false
+		for _, dir := range dirs {
+			nx, ny := ele.x+dir.x, ele.y+dir.y
+			if !(0 <= nx && nx < m && 0 <= ny && ny < n && grid[nx][ny] == originalColor) {
+				isBorder = true
+			} else if !vis[nx][ny] {
+				q = append(q, point{nx, ny})
+			}
+		}
+		if isBorder {
+			borders = append(borders, point{ele.x, ele.y})
+		}
+	}
+}
+```