1. 题目
Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.
For example,
Given n = 3,
You should return the following matrix:
[
[ 1, 2, 3 ],
[ 8, 9, 4 ],
[ 7, 6, 5 ]
]
2. 思路
是之前的Spiral Matrix的简化版本,https://segmentfault.com/a/11...
3. 代码
class Solution {
public:
// 一圈圈的转,每次转最外面的一圈。
vector<vector<int>> generateMatrix(int n) {
vector<vector<int>> matrix;
for (int i = 0; i < n; i++) {
matrix.push_back(vector<int>());
vector<int>& r = matrix.back();
for (int j = 0; j < n; j++) {
r.push_back(0);
}
}
int ki = 0;
int kj = 0; // 左上顶点的坐标
if (n == 0) {return matrix;}
int cnt = 1;
while (ki < (n+1)/2) {
// 最后单点
if (kj == n-kj-1) {
for (int i = ki; i < n-ki; i++) {
matrix[i][n-kj-1] = cnt++;
}
return matrix;
}
// 上行[ki, kj] -> [ki, n-kj-1]
for (int j = kj; j < n-kj-1; j++) {
matrix[ki][j] = cnt++;
}
// 右竖[ki, n-kj-1] -> [m-ki-1, n-kj-1]
for (int i = ki; i < n-ki-1; i++) {
matrix[i][n-kj-1] = cnt++;
}
// 下行[m-ki-1, kj] <- [m-ki-1, n-kj-1]
for (int j = n-kj-1; j > kj; j--) {
matrix[n-ki-1][j] = cnt++;
}
// 左竖[ki, kj] <- [m-ki-1, kj]
for (int i = n-ki-1; i > ki; i--) {
matrix[i][kj] = cnt++;
}
ki++;
kj++;
}
return matrix;
}
};
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