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Given a 2D board containing ‘X‘ and ‘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
题目大意：给一个棋盘，上面有‘X‘和‘O‘，找出所有被‘X‘包围的‘O‘，并替换成‘X‘。
解题思路：
解法一：从边开始，找出在四条边上的O，以BFS的方式找出所有与边上的O连通的O，这些O都是不被包围的，将这些O替换为某个特殊字符，遍历完之后，将O替换为X，将特殊字符替换为O。
解法二：直接用BFS遍历，找到一个O之后，加入BFS队列，找出与之连通的所有的O同时判断这些O是否被包围（即是否位于某条边上），将这些O加入一个List，每一个连通的O区域用一个surround布尔变量表示这个连通区域是否是被surround的，如果是就把这些O替换为X，否则不替换并继续遍历其他。
注意：这里因为要把i、j下标放入队列或者结果集，一开始我使用String类型key= i+"_"+j 来处理，后来看到别人key = i * colLen + j ; colLen是棋盘的列的数量，然后用key/colLen，key%colLen来得到i,j下标效率更高。
Talk is cheap>>
解法一：
public void solve(char[][] board) { Queue<Integer> queue = new ArrayDeque<>(); int rowLen = board.length; if (rowLen == 0) return; int colLen = board[0].length; int[][] adj = {{0, -1}, {0, 1}, {-1, 0}, {1, 0}}; for (int i = 0; i < rowLen; i++) { for (int j = 0; j < colLen; j++) {
//如果O在四条边上 if (board[i][j] == ‘O‘ && (i == 0 || j == 0 || i == rowLen - 1 || j == colLen - 1)) { int key = i * colLen + j; board[i][j] = ‘1‘; queue.add(key); while (!queue.isEmpty()) {
//BFS遍历与边上的O连通的O key = queue.poll(); int x = key / colLen; int y = key % colLen; for (int k = 0; k < 4; k++) { int adj_x = x + adj[k][0]; int adj_y = y + adj[k][1]; if (adj_x >= 0 && adj_y >= 0 && adj_x < rowLen && adj_y < colLen) { if (board[adj_x][adj_y] == ‘O‘) { int pos = adj_x * colLen + adj_y; board[adj_x][adj_y]=‘1‘; queue.add(pos); } } } } } } } for (int i = 0; i < rowLen; i++) { for (int j = 0; j < colLen; j++) { if (board[i][j] == ‘O‘) { board[i][j] = ‘X‘; } else if (board[i][j] == ‘1‘) { board[i][j] = ‘O‘; } } } }
解法二：
public void solve(char[][] board) { Queue<Integer> queue = new ArrayDeque<>(); int rowLen = board.length; if (rowLen == 0) { return; } int[][] adj = {{0, -1}, {0, 1}, {-1, 0}, {1, 0}}; int colLen = board[0].length; boolean[][] visited = new boolean[rowLen][colLen]; for (int i = 0; i < rowLen; i++) { for (int j = 0; j < colLen; j++) { if (!visited[i][j] && board[i][j] == ‘O‘) {
//标准的BFS遍历 boolean surround = true; List<Integer> toHandle = new ArrayList<>(); queue.add(i * colLen + j); while (!queue.isEmpty()) { int key = queue.poll(); toHandle.add(key); int x = key / colLen; int y = key % colLen; for (int k = 0; k < 4; k++) {
//检查O的上下左右 int adj_x = x + adj[k][0]; int adj_y = y + adj[k][1]; if (adj_x >= 0 && adj_y >= 0 && adj_x < rowLen && adj_y < colLen) {
//都不在边上 if (board[adj_x][adj_y] == ‘O‘ && !visited[adj_x][adj_y]) { int pos = adj_x * colLen + adj_y; queue.add(pos); visited[adj_x][adj_y] = true; } } else {
//有一个在边上，置surround=false surround = false; } } } if (surround) { for (int key : toHandle) { board[key / colLen][key % colLen] = ‘X‘; } } } } } }
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原文地址：http://www.cnblogs.com/aboutblank/p/4394511.html
