leet code 54 第二次做数组顺时针展开
2020-11-29 23:54 woshihuangrulin 阅读(103) 评论(0) 收藏 举报1, 将展开方向使用enum表示;
2,想好循环终止的条件-》转方向时下一次要访问的点越界或者已经访问过
class Solution {
public:
vector<int> spiralOrder(vector<vector<int>>& matrix) {
vector<int> result;
if (matrix.empty()) {
return result;
}
if (matrix.front().empty()) {
return result;
}
int m = matrix.size() - 1;
int n = matrix.front().size() - 1;
vector<vector<bool>> visted(matrix.size(), vector<bool>(matrix.front().size(), false));
int i = 0;
int j = 0;
enum Direction {RIGHT, DOWN, LEFT, UP};
Direction dir = Direction::RIGHT;
while(1) {
if (dir == Direction::RIGHT) {
result.emplace_back(matrix[i][j]);
visted[i][j] = true;
// 到达右边界或者到达上一轮访问过的位置
if (j == n || visted[i][j + 1] == true) {
if (i == m || visted[++i][j]) {
break;
}
dir = Direction::DOWN;
}
else if(visted[i][++j] == true) {
break;
}
}
else if (dir == Direction::DOWN) {
result.emplace_back(matrix[i][j]);
visted[i][j] = true;
if (i == m || visted[i + 1][j] == true) {
if (j == 0 || visted[i][--j]) {
break;
}
dir = Direction::LEFT;
}
else if(visted[++i][j] == true) {
break;
}
}
else if (dir == Direction::LEFT) {
result.emplace_back(matrix[i][j]);
visted[i][j] = true;
if (j == 0 || visted[i][j-1] == true) {
if (i == 0 || visted[--i][j]) {
break;
}
dir = Direction::UP;
}
else if (visted[i][--j] == true) {
break;
}
}
else {
result.emplace_back(matrix[i][j]);
visted[i][j] = true;
if (i == 0 || visted[i - 1][j] == true) {
if (j == n || visted[i][++j]) {
break;
}
dir = Direction::RIGHT;
}
else if (visted[--i][j] == true){
break;
}
}
}
return result;
}
};
浙公网安备 33010602011771号