Spiral Matrix
Given an m x n matrix, return all elements of the matrix in spiral order.
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Example 1:
Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [1,2,3,6,9,8,7,4,5]
Example 2:
Input: matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
Output: [1,2,3,4,8,12,11,10,9,5,6,7]
Constraints:
m == matrix.length
n == matrix[i].length
1 <= m, n <= 10
-100 <= matrix[i][j] <= 100
Solution
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class Solution {
public:
enum direction { right, down, left, up };
vector<int> spiralOrder(vector<vector<int>>& matrix) {
vector<int> ans;
vector<pair<int, int>> myVec;
set<pair<int, int>> mySet;
int row = matrix.size();
int col = matrix[0].size();
int n = row * col;
int i = 0, j = 0;
int d = right;
auto tmp = make_pair(0, 0);
mySet.insert(tmp);
myVec.push_back(tmp);
while (myVec.size() < n) {
if (d == right) {
int val = j + 1;
if (val < col && mySet.count(make_pair(i, val)) == 0) {
auto item = make_pair(i, val);
mySet.insert(item);
myVec.push_back(item);
j = val;
} else {
d = (d + 1) % 4;
}
} else if (d == down) {
int val = i + 1;
if (val < row && mySet.count(make_pair(val, j)) == 0) {
auto item = make_pair(val, j);
mySet.insert(item);
myVec.push_back(item);
i = val;
} else {
d = (d + 1) % 4;
}
} else if (d == left) {
int val = j - 1;
if (val >= 0 && mySet.count(make_pair(i, val)) == 0) {
auto item = make_pair(i, val);
mySet.insert(item);
myVec.push_back(item);
j = val;
} else {
d = (d + 1) % 4;
}
} else if (d == up) {
int val = i - 1;
if (val >= 0 && mySet.count(make_pair(val, j)) == 0) {
auto item = make_pair(val, j);
mySet.insert(item);
myVec.push_back(item);
i = val;
} else {
d = (d + 1) % 4;
}
}
}
for (auto item : myVec) {
ans.push_back(matrix[item.first][item.second]);
}
return ans;
}
};
