723. Candy Crush

O(mn) time
基本思路是先标记每行以及每列连续三个相同的candy，然后对每一列从下往上进行更新，最后反复更新board直到没有再需要调整的candy为止

class Solution {
public:
    vector<vector<int>> candyCrush(vector<vector<int>>& board) {
        int R = board.size(), C = board[0].size();
        bool changed = false;
        // 横向只要是3个相同就变成负数
        for (int r = 0; r < R; ++r) {
            for (int c = 0; c + 2 < C; ++c) {
                int v = abs(board[r][c]);
                if (v != 0 && v == abs(board[r][c + 1]) && v == abs(board[r][c + 2])) {
                    board[r][c] = board[r][c + 1] = board[r][c + 2] = -v;
                    changed = true;
                }
            }
        }
        // 纵向只要是3个相同就变成负数
        for (int r = 0; r + 2 < R; ++r) {
            for (int c = 0; c < C; ++c) {
                int v = abs(board[r][c]);
                if (v != 0 && v == abs(board[r + 1][c]) && v == abs(board[r + 2][c])) {
                    board[r][c] = board[r + 1][c] = board[r + 2][c] = -v;
                    changed = true;
                }
            }
        }
        if (!changed) return board;
        // 对于每一列从底往上移动
        for (int c = 0; c < C; ++c) {
            int i = R - 1; // 定义一个i表示当前要修改的元素
            for (int r = R - 1; r >= 0; --r) {
                if (board[r][c] > 0) {
                    board[i--][c] = board[r][c];
                }
            }
            for (; i >= 0; --i) { // 剩余元素全部置0
                board[i][c] = 0;
            }
        }
        return candyCrush(board);
    }
};