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126 lines
3.4 KiB
Markdown
Executable File
126 lines
3.4 KiB
Markdown
Executable File
# [130. Surrounded Regions](https://leetcode.com/problems/surrounded-regions/)
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## 题目
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Given a 2D board containing `'X'` and `'O'` (**the letter O**), capture all regions surrounded by `'X'`.
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A region is captured by flipping all `'O'`s into `'X'`s in that surrounded region.
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**Example**:
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X X X X
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X O O X
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X X O X
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X O X X
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After running your function, the board should be:
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X X X X
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X X X X
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X X X X
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X O X X
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**Explanation:**
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Surrounded regions shouldn’t be on the border, which means that any `'O'` on the border of the board are not flipped to `'X'`. Any `'O'` that is not on the border and it is not connected to an `'O'` on the border will be flipped to `'X'`. Two cells are connected if they are adjacent cells connected horizontally or vertically.
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## 题目大意
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给定一个二维的矩阵，包含 'X' 和 'O'（字母 O）。找到所有被 'X' 围绕的区域，并将这些区域里所有的 'O' 用 'X' 填充。被围绕的区间不会存在于边界上，换句话说，任何边界上的 'O' 都不会被填充为 'X'。 任何不在边界上，或不与边界上的 'O' 相连的 'O' 最终都会被填充为 'X'。如果两个元素在水平或垂直方向相邻，则称它们是“相连”的。
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## 解题思路
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- 给出一张二维地图，要求把地图上非边缘上的 'O' 都用 'X' 覆盖掉。
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- 这一题有多种解法。第一种解法是并查集。先将边缘上的 'O' 全部都和一个特殊的点进行 `union()` 。然后再把地图中间的 'O' 都进行 `union()`，最后把和特殊点不是同一个集合的点都标记成 'X'。第二种解法是 DFS 或者 BFS，可以先将边缘上的 'O' 先标记成另外一个字符，然后在递归遍历过程中，把剩下的 'O' 都标记成 'X'。
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## 代码
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```go
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package leetcode
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import (
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"github.com/halfrost/LeetCode-Go/template"
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)
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// 解法一 并查集
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func solve(board [][]byte) {
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if len(board) == 0 {
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return
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}
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m, n := len(board[0]), len(board)
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uf := template.UnionFind{}
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uf.Init(n*m + 1) // 特意多一个特殊点用来标记
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for i := 0; i < n; i++ {
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for j := 0; j < m; j++ {
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if (i == 0 || i == n-1 || j == 0 || j == m-1) && board[i][j] == 'O' { //棋盘边缘上的 'O' 点
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uf.Union(i*m+j, n*m)
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} else if board[i][j] == 'O' { //棋盘非边缘上的内部的 'O' 点
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if board[i-1][j] == 'O' {
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uf.Union(i*m+j, (i-1)*m+j)
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}
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if board[i+1][j] == 'O' {
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uf.Union(i*m+j, (i+1)*m+j)
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}
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if board[i][j-1] == 'O' {
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uf.Union(i*m+j, i*m+j-1)
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}
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if board[i][j+1] == 'O' {
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uf.Union(i*m+j, i*m+j+1)
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}
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}
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}
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}
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for i := 0; i < n; i++ {
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for j := 0; j < m; j++ {
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if uf.Find(i*m+j) != uf.Find(n*m) {
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board[i][j] = 'X'
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}
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}
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}
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}
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// 解法二 DFS
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func solve1(board [][]byte) {
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for i := range board {
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for j := range board[i] {
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if i == 0 || i == len(board)-1 || j == 0 || j == len(board[i])-1 {
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if board[i][j] == 'O' {
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dfs130(i, j, board)
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}
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}
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}
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}
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for i := range board {
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for j := range board[i] {
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if board[i][j] == '*' {
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board[i][j] = 'O'
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} else if board[i][j] == 'O' {
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board[i][j] = 'X'
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}
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}
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}
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}
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func dfs130(i, j int, board [][]byte) {
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if i < 0 || i > len(board)-1 || j < 0 || j > len(board[i])-1 {
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return
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}
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if board[i][j] == 'O' {
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board[i][j] = '*'
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for k := 0; k < 4; k++ {
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dfs130(i+dir[k][0], j+dir[k][1], board)
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}
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}
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}
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``` |