130. Surrounded Regions
阿新 • • 發佈:2017-07-23
ons void while boa apt pty 改變 int style
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
矩陣的bfs, 套路一致,從外向內, 很easy
構造類, 遍歷矩陣建立圖
bfs要點在於如何建圖, 是否建類, 建比較器, 建方向容器, 建走過的路的存儲器(數組, 或者set, list) 如何遍歷(堆不空?), 遍歷到內部的點時判斷是否符合題意(邊界, 走過), 再考慮題意, 判斷當前的點是否符合題意來加入結果的容器 或結果值, 將當前的點加入堆中是否要改變值啊什麽的, 關鍵在於建立的是什麽圖. 裏面的點是怎麽個意思, 都是題意的轉化
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‘; } } } }
130. Surrounded Regions