好傻逼。。这个题麻烦死了。
简单的说，就是里面的O能想办法连到外面的O就不用被转换，否则就变为X。
需要2个额外的BOARD 一个记录VISIT 一个记录最后剩下的O
简单的说就是从最外面那圈开始，如果是O，就作为一个起点往上下左右找
很多情况都不用再找了： 找到X，到边（除了一开始），VISIT过
不是上述情况，就当前VisiT=TRUE 以免又循环回来，然后再四面八方。
检测是不是边，要么专门设个FLAG，如果是一开始就不检测，不是一开始就检测。
总之很麻烦，很容易就OVERFLOW
public void solve(char[][] board)
{
int row = board.length;
if(row == 0) return;
int col = board[0].length;
char[][] check = new char[row][col];
boolean[][] visit = new boolean[row][col];
//m = 0;
for(int n = 0; n < col; n++)
{
if(board[0][n] == 'O')
{
visit = new boolean[row][col];
helper(0,n,board,check,visit,true);
}
if(board[board.length-1][n] == 'O')
{
visit = new boolean[row][col];
helper(board.length-1,n,board,check,visit,true);
}
}
for(int m = 1; m < row-1; m++)
{
if(board[m][0] == 'O')
{
visit = new boolean[row][col];
helper(m,0,board,check,visit,true);
}
if(board[m][board[0].length-1] == 'O')
{
visit = new boolean[row][col];
helper(m,board[0].length-1,board,check,visit,true);
}
}
// System.out.println(board[1][1]);
for(int m = 1; m < row-1;m++)
for(int n = 1; n < col-1;n++)
{
if(board[m][n] == 'X')
{
}
else
{
if(check[m][n] == 'O')
{
board[m][n] = 'O';
}
else board[m][n] = 'X';
}
}
}
public void helper(int m,int n,char[][] board, char[][] check, boolean[][] visit,boolean init)
{
if(check[m][n] == 'X' || visit[m][n]) return;
if((!init) && (m == 0 || n == 0 || m == board.length-1|| n == board[0].length-1)) return;
//System.out.println(m + " " +n);
check[m][n] = 'O';
visit[m][n] = true;
if(m>0 && board[m-1][n] == 'O') helper(m-1,n,board,check,visit,false);
if(n>0 && board[m][n-1] == 'O') helper(m,n-1,board,check,visit,false);
if(m < board.length-1 && board[m+1][n] == 'O') helper(m+1,n,board,check,visit,false);
if(n < board[0].length-1 && board[m][n+1] == 'O') helper(m,n+1,board,check,visit,false);
}
搞了好久。。。
被自己蠢哭了。
visit[i][j] = true写成了visit[j][j] = true 我DEBUG了1个多小时。。。。。。。
image
从最外面那圈的所有O开始顺着相邻的O走，里面那些能走到的O就是不用变成反革命分子的。否则最后变成X，反革命分子。
主要是剪枝，乱七八糟一大堆情况，包括但不限于：
走到边了，走出界了，已经接受过组织的审查了， 他本身就是反革命分子，etc..
一开始4*4 test case没问题让我以为代码是正确的。结果后来各种stackoverflow，以为是不够好，改了依然如此，绝望中浪费了整整他妈的一个多小时。
public class Solution
{
public void solve(char[][] board)
{
if(board.length == 0) return;
boolean[][] moveVisit = new boolean[board.length][board[0].length];
boolean[][] ok = new boolean[board.length][board[0].length];
for(int i = 0; i < board.length;i++)
{
if(board[i][0] == 'O')
{
go(board,moveVisit,ok,i,0,true);
moveVisit = new boolean[board.length][board[0].length];
}
if(board[i][board[0].length-1] == 'O')
{
go(board,moveVisit,ok,i,board[0].length-1,true);
moveVisit = new boolean[board.length][board[0].length];
}
}
for(int i = 1; i < board[0].length;i++)
{
if(board[0][i] == 'O')
{
go(board,moveVisit,ok,0,i,true);
moveVisit = new boolean[board.length][board[0].length];
}
if(board[board.length-1][i]== 'O')
{
go(board,moveVisit,ok,board.length-1,i,true);
}
}
for(int i = 1; i < board.length-1;i++)
{
for(int j = 1; j < board[0].length-1;j++)
{
if(board[i][j] == 'X') continue;
if(!ok[i][j]) board[i][j] = 'X';
}
}
return;
}
public void go(char[][] board, boolean[][] moveVisit, boolean[][] ok, int i, int j,boolean begin)
{
if(board[i][j] == 'X' || moveVisit[i][j]) return;
if(!begin && (i == 0 || j == 0 || i == board.length-1 || j == board[0].length -1)) return;
ok[i][j] = true;
moveVisit[i][j] = true;
if(i > 0 && board[i-1][j] == 'O') go(board,moveVisit,ok,i-1,j,false);
if(j>0 && board[i][j-1] == 'O') go(board,moveVisit,ok,i,j-1,false);
if(i < board.length-1 && board[i+1][j] == 'O') go(board,moveVisit,ok,i+1,j,false);
if(j < board[0].length-1 && board[i][j+1] == 'O') go(board,moveVisit,ok,i,j+1,false);
return;
}
}
三刷。
刚养成DFS开始CHECK的习惯，这个题瞬间打脸。。
这个题不但开始要CHECK，最完美的情况是进入前再CHECK一下，保证不是最外面那圈，再进去……
Time: O(mn)
Space: O(1) 我好像一直把空间复杂度混淆了，我前面所指的空间复杂度其实指的是extra space...
public class Solution {
public void solve(char[][] board) {
if (board.length == 0) return;
int row = board.length;
int col = board[0].length;
for (int i = 0; i < row; i++) {
if (board[i][0] == 'O')
dfs(board, i, 0);
if (board[i][col-1] == 'O')
dfs(board, i, col-1);
}
for (int i = 1; i < col - 1; i++) {
if (board[0][i] == 'O')
dfs(board, 0, i);
if (board[row - 1][i] == 'O')
dfs(board,row-1, i);
}
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
if (board[i][j] == '1') {
board[i][j] = 'O';
} else {
board[i][j] = 'X';
}
}
}
return;
}
public void dfs(char[][] board, int m, int n) {
if (board[m][n] != 'O') return;
board[m][n] = '1';
if (m < board.length - 2)
dfs(board, m + 1, n);
if (m > 1)
dfs(board, m - 1, n);
if (n < board[0].length - 2)
dfs(board, m, n + 1);
if (n > 1)
dfs(board, m, n - 1);
}
}