[LeetCode][JavaScript]Surrounded Regions

Surrounded Regions

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

https://leetcode.com/problems/surrounded-regions/

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想通了其实很简单，但我就是折腾了好久...

第一轮找外围所有的O，这些点最后肯定不会被替换成X。

然后进行dfs，与这些点相连的点同样不会是X。

把这些不可能是X的点都标记成S，最后把所有O替换成X，把所有S替换成O。

我一开始尝试正着做，先缓存下需要标记的点，一口气都标成X，妥妥的MLE。

第二次尝试用bfs，先标记成X，如果不行回溯回去改成O，递归太深堆栈爆了...真是拙计

/**
 * @param {character[][]} board
 * @return {void} Do not return anything, modify board in-place instead.
 */
var solve = function(board) {
    var m = board.length, n, i, j, queue = [];
    for(i = 0; i < m; i++){
        n = board[0].length;
        for(j = 0; j < n; j++){
            if(board[i][j] === 'O' && (i === 0 || i === m - 1 || j === 0 || j === n - 1)){
                queue.push({row : i, col : j});
            }
           }
    }
    bfs(queue);
    for(i = 0; i < m; i ++){
        for(j = 0; j < n; j ++){
            if(board[i][j] === 'O'){
                board[i][j] = 'X';
            }else if(board[i][j] === 'S'){
                board[i][j] = 'O';
            }
        }
    }

    function bfs(queue){
        var i, j;
        while(queue.length > 0){
            var top = queue.pop();
            i = top.row, j = top.col;
            if(board[i][j] === 'S'){
                continue;
            }
            board[i][j] = 'S';
            if(board[i + 1] && board[i + 1][j] && board[i + 1][j] === 'O'){
                queue.push({row : i + 1, col : j});
            }
            if(board[i - 1] && board[i - 1][j] && board[i - 1][j] === 'O'){
                queue.push({row : i - 1, col : j});
            }
            if(board[i][j + 1] && board[i][j + 1] === 'O'){
                queue.push({row : i, col : j + 1});
            }
            if(board[i][j - 1] && board[i][j - 1] === 'O'){
                queue.push({row : i, col : j - 1});
            }
        }
    }
};

TLE：

/**
 * @param {character[][]} board
 * @return {void} Do not return anything, modify board in-place instead.
 */
var solveTLE = function(board) {
    var m = board.length, n, i, j;
    for(i = 0; i < m; i++){
        n = board[0].length;
        for(j = 0; j < n; j++){
            if(board[i][j] === 'O'){
                bfs([{row : i, col : j}]);
            }
           }
    }
    for(i = 0; i < m; i++){
        for(j = 0; j < n; j++){
            if(board[i][j] === 'A'){
                board[i][j] = 'X';
            }
        }
    }

    function bfs(queue){
        var len = queue.length, top, i, j, res;
        while(len--){
            top = queue.pop();        
            i = top.row; j = top.col;
            if(!board[i] || !board[i][j]){
                return false;
            }
            if(board[i][j] === 'O'){                
                queue.push({row : i + 1, col : j});
                queue.push({row : i - 1, col : j});
                queue.push({row : i, col : j + 1});
                queue.push({row : i, col : j - 1});
            }
        }
        if(queue.length !== 0){
            board[i][j] = 'A';
            res = bfs(queue);
            if(!res){
                board[i][j] = 'O';
                return false;
            }
        }
        return true;
    }
};

MLE写得太丑不贴了...