标签：数组 region logs log oid ons boolean cti blog
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. For 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
矩阵的bfs, 套路一致,从外向内, 很easy
构造类, 遍历矩阵建立图
bfs要点在于如何建图, 是否建类, 建比较器, 建方向容器, 建走过的路的存储器(数组, 或者set, list) 如何遍历(堆不空?), 遍历到内部的点时判断是否符合题意(边界, 走过), 再考虑题意, 判断当前的点是否符合题意来加入结果的容器 或结果值, 将当前的点加入堆中是否要改变值啊什么的, 关键在于建立的是什么图. 里面的点是怎么个意思, 都是题意的转化
class Cell {
int x, y;
public Cell(int x, int y) {
this.x = x;
this.y = y;
}
}
public class Solution {
public void solve(char[][] board) {
if(board == null || board.length == 0 || board[0].length == 0) return;
int m = board.length;
int n = board[0].length;
Queue<Cell> q = new LinkedList<>();
int[][] directions = new int[][]{{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
boolean [][] visited = new boolean[m][n];
for (int j=0; j<board[0].length; j++) {
if (board[0][j] == ‘O‘) q.offer(new Cell(0, j));
if (board[board.length-1][j] == ‘O‘) q.offer(new Cell(board.length-1, j));
visited[0][j] = true;
visited[m - 1][j] = true;
}
for (int i=0; i<board.length; i++) {
if (board[i][0] == ‘O‘) q.offer(new Cell(i, 0));
if (board[i][board[0].length-1] == ‘O‘) q.offer(new Cell(i, board[0].length-1));
visited[i][0] = true;
visited[i][n - 1] = true;
}
while (!q.isEmpty()) {
Cell cur = q.poll();
board[cur.x][cur.y] = ‘$‘;
for (int[] dir : directions) {
int x = dir[0] + cur.x;
int y = dir[1] + cur.y;
if (x < 0 || y < 0 || x >= m || y >= n || visited[x][y] || board[x][y] == ‘X‘) {
continue;
}
visited[x][y] = true;
q.offer(new Cell(x, y));
}
}
for (int i=0; i<board.length; i++) {
for (int j=0; j<board[0].length; j++) {
if (board[i][j] == ‘X‘) continue;
else if (board[i][j] == ‘$‘) board[i][j] = ‘O‘;
else if (board[i][j] == ‘O‘) board[i][j] = ‘X‘;
}
}
}
}
标签：数组 region logs log oid ons boolean cti blog
原文地址：http://www.cnblogs.com/apanda009/p/7226100.html
