题目描述:
给定一个包含 m x n 个元素的矩阵(m 行, n 列),请按照顺时针螺旋顺序,返回矩阵中的所有元素。
示例 1:
输入:
[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]
输出: [1,2,3,6,9,8,7,4,5]
示例 2:
输入:
[
[1, 2, 3, 4],
[5, 6, 7, 8],
[9,10,11,12]
]
输出: [1,2,3,4,8,12,11,10,9,5,6,7]
解法:
class Solution {
public:
vector<int> spiralOrder(vector<vector<int>>& matrix) {
int x = 0, y = 0;
bool right = true, down = false, left = false, up = false;
int dx = 0, dy = 1;
vector<int> res;
int m = matrix.size();
if(m == 0){
return res;
}else if(m == 1){
return matrix[0];
}
int n = matrix[0].size();
if(n == 0){
return res;
}else if(n == 1){
for(vector<int> lst : matrix){
res.push_back(lst[0]);
}
return res;
}
int idx = m*n;
int min_x = 1, max_x = m-1;
int min_y = 0, max_y = n-1;
while(idx--){
res.push_back(matrix[x][y]);
x += dx;
y += dy;
if(right && y == max_y){
right = false;
down = true;
dx = 1;
dy = 0;
max_y--;
}else if(down && x == max_x){
down = false;
left = true;
dx = 0;
dy = -1;
max_x--;
}else if(left && y == min_y){
left = false;
up = true;
dx = -1;
dy = 0;
min_y++;
}else if(up && x == min_x){
up = false;
right = true;
dx = 0;
dy = 1;
min_x++;
}
}
return res;
}
};