Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.
For example,
Given the following matrix:
Given the following matrix:
[ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ]
You should return
[1,2,3,6,9,8,7,4,5]
.
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class Solution { | |
public: | |
vector<int> spiralOrder(vector<vector<int>>& matrix) { | |
vector<int> result; | |
if(matrix.empty()) | |
return result; | |
int m=matrix.size(); | |
int n=matrix[0].size(); | |
int u=n, r=m, d=n, l=m; | |
int row=0, col=0; | |
while(true){ | |
if(u==0) | |
return result; | |
int t=col+u; | |
for(;col<t; col++){ | |
result.push_back(matrix[row][col]); | |
} | |
col--; r--; l--; row++; | |
if(r==0) | |
return result; | |
t=row+r; | |
for(;row<t; row++){ | |
result.push_back(matrix[row][col]); | |
} | |
row--; d--; u--; col--; | |
if(d==0) | |
return result; | |
t=col-(d-1); | |
for(;col>=t; col--){ | |
result.push_back(matrix[row][col]); | |
} | |
col++; l--; r--; row--; | |
if(l==0) | |
return result; | |
t=row-(l-1); | |
for(;row>=t; row--){ | |
result.push_back(matrix[row][col]); | |
} | |
row++; u--; d--; col++; | |
} | |
} | |
}; |
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