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LeetCode-130

题目

Snipaste_2020-11-19_13-04-50.png

结果

Snipaste_2020-11-19_13-03-53.png

代码

如果一个‘O’没有被包围,一定是因为他在边界或者他跟边界的‘O’有勾连,所以边界的‘O’是始作俑者,罪魁祸首。所以从边界的‘O’出发,把与其有勾连的全部标记一遍,就知道board里的每个‘O’是什么成分。

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class Solution {

public void solve(char[][] board) {
if (board.length == 0) {
return;
}

boolean[][] flag = new boolean[board.length][board[0].length];
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
if (board[i][j] == 'O' && onBoundary(board, i, j) && !flag[i][j]) {
flag[i][j] = true;
spread(board, flag, i, j);
}
}
}
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
if (board[i][j] == 'O' && !flag[i][j]) {
board[i][j] = 'X';
}
}
}
}

private void spread(char[][] board, boolean[][] flag, int row, int col) {
if (row > 0 && board[row - 1][col] == 'O' && !flag[row - 1][col]) {
flag[row - 1][col] = true;
spread(board, flag, row - 1, col);
}
if (row < board.length - 1 && board[row + 1][col] == 'O' && !flag[row + 1][col]) {
flag[row + 1][col] = true;
spread(board, flag, row + 1, col);
}
if (col > 0 && board[row][col - 1] == 'O' && !flag[row][col - 1]) {
flag[row][col - 1] = true;
spread(board, flag, row, col - 1);
}
if (col < board[0].length - 1 && board[row][col + 1] == 'O' && !flag[row][col + 1]) {
flag[row][col + 1] = true;
spread(board, flag, row, col + 1);
}
}

private boolean onBoundary(char[][] board, int row, int col) {
return row == 0 || row == board.length - 1 || col == 0 || col == board[0].length - 1;
}
}

复杂度

时间复杂度:O(n)

空间复杂度:O(n),n为元素个数row×col。