0130. Surrounded Regions

# 130. Surrounded Regions#

## 题目 #

Given a 2D board containing `'X'` and `'O'` (the letter O), capture all regions surrounded by `'X'`.

A region is captured by flipping all `'O'`s into `'X'`s in that surrounded region.

Example:

``````X X X X
X O O X
X X O X
X O X X
``````

After running your function, the board should be:

``````X X X X
X X X X
X X X X
X O X X
``````

Explanation:

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.

## 解题思路 #

• 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。
• 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’。

## 代码 #

``````
package leetcode

import (
"github.com/halfrost/leetcode-go/template"
)

// 解法一 并查集
func solve(board [][]byte) {
if len(board) == 0 {
return
}
m, n := len(board[0]), len(board)
uf := template.UnionFind{}
uf.Init(n*m + 1) // 特意多一个特殊点用来标记

for i := 0; i < n; i++ {
for j := 0; j < m; j++ {
if (i == 0 || i == n-1 || j == 0 || j == m-1) && board[i][j] == 'O' { //棋盘边缘上的 'O' 点
uf.Union(i*m+j, n*m)
} else if board[i][j] == 'O' { //棋盘非边缘上的内部的 'O' 点
if board[i-1][j] == 'O' {
uf.Union(i*m+j, (i-1)*m+j)
}
if board[i+1][j] == 'O' {
uf.Union(i*m+j, (i+1)*m+j)
}
if board[i][j-1] == 'O' {
uf.Union(i*m+j, i*m+j-1)
}
if board[i][j+1] == 'O' {
uf.Union(i*m+j, i*m+j+1)
}

}
}
}
for i := 0; i < n; i++ {
for j := 0; j < m; j++ {
if uf.Find(i*m+j) != uf.Find(n*m) {
board[i][j] = 'X'
}
}
}
}

// 解法二 DFS
func solve1(board [][]byte) {
for i := range board {
for j := range board[i] {
if i == 0 || i == len(board)-1 || j == 0 || j == len(board[i])-1 {
if board[i][j] == 'O' {
dfs130(i, j, board)
}
}
}
}

for i := range board {
for j := range board[i] {
if board[i][j] == '*' {
board[i][j] = 'O'
} else if board[i][j] == 'O' {
board[i][j] = 'X'
}
}
}
}

func dfs130(i, j int, board [][]byte) {
if i < 0 || i > len(board)-1 || j < 0 || j > len(board[i])-1 {
return
}
if board[i][j] == 'O' {
board[i][j] = '*'
for k := 0; k < 4; k++ {
dfs130(i+dir[k][0], j+dir[k][1], board)
}
}
}

``````

Apr 8, 2023