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[LeetCode][JavaScript]Surrounded Regions

分类: IT文章 • 2022-08-29 20:32:31

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/

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

 1 /**
 2  * @param {character[][]} board
 3  * @return {void} Do not return anything, modify board in-place instead.
 4  */
 5 var solve = function(board) {
 6     var m = board.length, n, i, j, queue = [];
 7     for(i = 0; i < m; i++){
 8         n = board[0].length;
 9         for(j = 0; j < n; j++){
10             if(board[i][j] === 'O' && (i === 0 || i === m - 1 || j === 0 || j === n - 1)){
11                 queue.push({row : i, col : j});
12             }
13            }
14     }
15     bfs(queue);
16     for(i = 0; i < m; i ++){
17         for(j = 0; j < n; j ++){
18             if(board[i][j] === 'O'){
19                 board[i][j] = 'X';
20             }else if(board[i][j] === 'S'){
21                 board[i][j] = 'O';
22             }
23         }
24     }
25 
26     function bfs(queue){
27         var i, j;
28         while(queue.length > 0){
29             var top = queue.pop();
30             i = top.row, j = top.col;
31             if(board[i][j] === 'S'){
32                 continue;
33             }
34             board[i][j] = 'S';
35             if(board[i + 1] && board[i + 1][j] && board[i + 1][j] === 'O'){
36                 queue.push({row : i + 1, col : j});
37             }
38             if(board[i - 1] && board[i - 1][j] && board[i - 1][j] === 'O'){
39                 queue.push({row : i - 1, col : j});
40             }
41             if(board[i][j + 1] && board[i][j + 1] === 'O'){
42                 queue.push({row : i, col : j + 1});
43             }
44             if(board[i][j - 1] && board[i][j - 1] === 'O'){
45                 queue.push({row : i, col : j - 1});
46             }
47         }
48     }
49 };

TLE：

 1 /**
 2  * @param {character[][]} board
 3  * @return {void} Do not return anything, modify board in-place instead.
 4  */
 5 var solveTLE = function(board) {
 6     var m = board.length, n, i, j;
 7     for(i = 0; i < m; i++){
 8         n = board[0].length;
 9         for(j = 0; j < n; j++){
10             if(board[i][j] === 'O'){
11                 bfs([{row : i, col : j}]);
12             }
13            }
14     }
15     for(i = 0; i < m; i++){
16         for(j = 0; j < n; j++){
17             if(board[i][j] === 'A'){
18                 board[i][j] = 'X';
19             }
20         }
21     }
22 
23     function bfs(queue){
24         var len = queue.length, top, i, j, res;
25         while(len--){
26             top = queue.pop();        
27             i = top.row; j = top.col;
28             if(!board[i] || !board[i][j]){
29                 return false;
30             }
31             if(board[i][j] === 'O'){                
32                 queue.push({row : i + 1, col : j});
33                 queue.push({row : i - 1, col : j});
34                 queue.push({row : i, col : j + 1});
35                 queue.push({row : i, col : j - 1});
36             }
37         }
38         if(queue.length !== 0){
39             board[i][j] = 'A';
40             res = bfs(queue);
41             if(!res){
42                 board[i][j] = 'O';
43                 return false;
44             }
45         }
46         return true;
47     }
48 };

MLE写得太丑不贴了...

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