[leetcode]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 XX O O XX X O XX O X X

After running your function, the board should be:

X X X XX X X XX X X XX O X X

题目的意思就是将所有被x包围的o全部替换为x，边界上的除外

思想

1、首先将第一行，第一列，最后一行，最后一列空出来

2、第一次遍历自左上至右下，如果当前元素的前一个元素与其正上方的元素为x则本身至x，否则不变， 再从右至左的做一遍

3、第二遍遍历自右下至左上，每行中的元素要从左向右再从右向左遍历一遍

4、第三遍遍历自左上至右下，同3，这遍是校验的，因为之前总是有点小错误，始终会出现其中有一行的一个元素出点错，所以就在做了一遍，由于不方便调试所以就这么做了，a过了的

void solve(vector<vector<char> > &board) {int m = board.size();if(m == 0) return;       int n = board[0].size();if(n == 0) return;if(m == 1 || n == 1) return;       vector<vector<bool> > vvc(m, vector<bool>(n, false));char x = 'X', o = 'O';        //根据前一个元素与上一个元素来置o for(int i = 1; i < m - 1; i++){ //从第二行开始的for(int j = 1; j < n - 1; j++){ //从第二列if(board[i][j] == o){vvc[i][j] = true;if(board[i - 1][j] == x &&  board[i][j - 1] == x){ board[i][j] = x; }}}for(int j = n - 2; j > 0; j--){ //从右向左再做一遍还原本来应该为o的元素   倒数第二列if(vvc[i][j]){if(board[i][j + 1] == o || board[i - 1][j] == o){ board[i][j] = o; }}}}        //从右下到左上for(int i = m - 2; i >= 1; i--){for(int j = n - 2; j >= 1; j--){ //从右向左if(vvc[i][j]){if(board[i + 1][j] == o  || board[i][j + 1] == o){  board[i][j] = o; }}}for(int j = 1; j < n - 1; j++){ //从左向右if(vvc[i][j]){if(board[i + 1][j] == o  || board[i][j - 1] == o){ board[i][j] = o; }}}}for(int i = 1; i < m - 1; i++){for(int j = 1; j < n - 1; j++){if(vvc[i][j]){if(board[i - 1][j] == o ||  board[i][j - 1] == o){ board[i][j] = o; }}}for(int j = n - 2; j > 0; j--){if(vvc[i][j]){if(board[i][j + 1] == o || board[i - 1][j] == o){board[i][j] = o;}}}}}