1 题目描述
给定一个包含字母’O’与’X’的二维平面,占领所有被’X’包围的区域(将被’X’包围的区域中的’O’填充为’X’即为占领)。
例子:
输入:
X X X X
X O O X
X X O X
X O X X
执行程序后应为:
X X X X
X X X X
X X X X
X O X X
注:
包围的区域不应在边界上,即任何在边界上的’O’不应被替换为’X’;若两个含’O’的单元水平方向或垂直方向连接即认为两部分是连接的;因此,该题目需将任何不在边界上并且不与边界上的’O’连接的’O’替换为’X’。
题目出处:
https://leetcode.com/problems/surrounded-regions/
2 解决思路
找出该平面左、下、右、上边界上所有的’O’及其坐标,遍历这些边界上的’O’,由其出发寻找所有与其直接以及间接连通的’O’并记录下来,最后除这些连通的’O’以外,将平面上所有其他的‘O’置为‘X’。
3 golang实现代码
https://github.com/leileiluoluo/leetcode/blob/master/130_Surrounded_Regions/test.go
// findOBoundaries return coordinates that value is 'O'
// and mark the coordinates visited
func findOBoundaries(board [][]byte, oVisited *[][]int) [][]int {
var boundaries [][]int
if 1 == len(board) {
if 'O' == board[0][0] {
p := []int{0, 0}
boundaries = append(boundaries, p)
*oVisited = append(*oVisited, p)
}
return boundaries
}
// left boundary
for i := 0; i < len(board)-1; i++ {
if 'O' == board[i][0] {
p := []int{i, 0}
boundaries = append(boundaries, p)
*oVisited = append(*oVisited, p)
}
}
// bottom boundary
for j := 0; j < len(board[0])-1; j++ {
if 'O' == board[len(board)-1][j] {
p := []int{len(board) - 1, j}
boundaries = append(boundaries, p)
*oVisited = append(*oVisited, p)
}
}
// right boundary
for i := len(board) - 1; i > 0; i-- {
if 'O' == board[i][len(board[0])-1] {
p := []int{i, len(board[0]) - 1}
boundaries = append(boundaries, p)
*oVisited = append(*oVisited, p)
}
}
// top boundary
for j := len(board[0]) - 1; j > 0; j-- {
if 'O' == board[0][j] {
p := []int{0, j}
boundaries = append(boundaries, p)
*oVisited = append(*oVisited, p)
}
}
return boundaries
}
func isVisited(i, j int, oVisited *[][]int) bool {
for _, v := range *oVisited {
if v[0] == i && v[1] == j {
return true
}
}
return false
}
func findOsAround(board [][]byte, i, j int, oVisited *[][]int) [][]int {
var osAround [][]int
// left
if j > 0 {
left := board[i][j-1]
if !isVisited(i, j-1, oVisited) && 'O' == left {
p := []int{i, j - 1}
osAround = append(osAround, p)
*oVisited = append(*oVisited, p)
}
}
// down
if i < len(board)-1 {
down := board[i+1][j]
if !isVisited(i+1, j, oVisited) && 'O' == down {
p := []int{i + 1, j}
osAround = append(osAround, p)
*oVisited = append(*oVisited, p)
}
}
// right
if j < len(board[0])-1 {
right := board[i][j+1]
if !isVisited(i, j+1, oVisited) && 'O' == right {
p := []int{i, j + 1}
osAround = append(osAround, p)
*oVisited = append(*oVisited, p)
}
}
// up
if i > 0 {
up := board[i-1][j]
if !isVisited(i-1, j, oVisited) && 'O' == up {
p := []int{i - 1, j}
osAround = append(osAround, p)
*oVisited = append(*oVisited, p)
}
}
return osAround
}
func findOs(board [][]byte, points [][]int, oVisited *[][]int) [][]int {
var os [][]int
for _, p := range points {
os = append(os, findOsAround(board, p[0], p[1], oVisited)...)
}
return os
}
func solve(board [][]byte) {
if 0 == len(board) {
return
}
var oVisited [][]int
// visit all 'O's from boundaries that is 'O'
os := findOBoundaries(board, &oVisited)
for {
os = findOs(board, os, &oVisited)
if len(os) == 0 {
break
}
}
// alter the not visited 'O' to 'X'
for i := range board {
for j := range board[i] {
if 'O' == board[i][j] && !isVisited(i, j, &oVisited) {
board[i][j] = 'X'
}
}
}
}