Spiral Matrix

难度：1

1Y又失败。。

解法：模拟。。

class Solution {public:const int LEFT=0;const int RIGHT=1;const int DOWN=2;const int UP=3;    vector<int> spiralOrder(vector<vector<int> > &matrix) {vector<int>ans;if(matrix.size() == 0)return ans;if(matrix.size() == 1)return matrix[0];if(matrix[0].size() == 0)return ans;int m=matrix.size();//rowint n=matrix[0].size();//colint cur_x=0,cur_y=0,cur_dir=RIGHT;ans.push_back(matrix[cur_x][cur_y]);int border_left=0;int border_right=n-1;int border_down=m-1;int border_up=0;while(border_left<=border_right&&border_up<=border_down){if(cur_x == border_up && cur_x == border_down && cur_y == border_left && cur_y == border_right){break;}int next_x,next_y;if(cur_dir == LEFT){next_x=cur_x;next_y=cur_y-1;if(next_y < border_left){cur_dir=UP;border_down--;continue;}}else if(cur_dir == RIGHT){next_x=cur_x;next_y=cur_y+1;if(next_y>border_right){cur_dir=DOWN;border_up++;continue;}}else if(cur_dir == DOWN){next_x=cur_x+1;next_y=cur_y;if(next_x>border_down){cur_dir=LEFT;border_right--;continue;}}else{next_x=cur_x-1;next_y=cur_y;if(next_x<border_up){cur_dir=RIGHT;border_left++;continue;}}ans.push_back(matrix[next_x][next_y]);cur_x=next_x;cur_y=next_y;}return ans;    }};